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Unit 5 : Kirchhoff ’s Laws and Electric Circuits

  • Unit 5 : Kirchhoff ’s Laws and Electric Circuits

    Key Unit Competence
    Analyse the complex electric circuits using Kirchhoff’s laws 
    My goals
    By the end of this unit, I will be able to:
    * analyse complex electric circuits using Kirchhoff’s laws.
    * identify sources of electric current.
    * describe components of simple electric circuits. 
    * state Kirchhoff’s laws and apply them to solve problems in electric 
    circuits.
    * acquire practical skills to manipulate apparatus and evaluate 
    experimental producers.
    * explain the differences between the potential difference and 
    electromotive forces.
    Introductory activity 
    Provided two bulbs, cell holder, bulb holder, voltmeter, ammeter, connecting 
    wires and a switch.
    C
    1. Make a simple electric circuit (as in the above figure) and comment 
    on your observation
    2. On the circuit made in (1), add a voltmeter and an ammeter.
    3. What would happen if you remove one bulb from the circuit?
    4. Would there be another type of connection of the circuit? Try it and 
    comment your observation.
    5. Discuss on the function of each element in provided list.
    Introduction
    This unit is one of the most interesting units in Physics. Even if you 
    ask someone who did not have enough studies in Physics he or she 
    will tell you that People studying physics will be engineers specifically 
    electricians. This Unit addresses the principles those electricians use in 
    their career.
    Review of elements of simple electric circuits 
    and their respective role
    An electric current consists of moving electric charges. Electric current must 
    flow in electric devices connected by conductors (wires). The motion of 
    electrons in a conductor is compared to water flow in a pipe. To move electrons, 
    there must be a source of electric current, a cell, a battery, a generator which 
    acts as a pump of water.
    Making a simple circuit
    Activity 1 
    Making a simple electric circuit with a bulb, a battery and wires
    Materials:
    * 2 pieces of copper wire
    * 1 bulb
    * 1 battery
    Procedure
    1. Examine diagrams A-J below. Predict whether the circuit will be 
    complete, and record your prediction on the chart below.
    2. Demonstrate the arrangements to test your predictions..
    PREDICTION CHART
    C
    Activity 1
    C
    What makes the bulb light?
    You may already understand an electrical circuit, or this may seem 
    like magic to you. Give what your teacher demonstrated some thought. 
    Why do you think the bulb in the diagram lights?
    C
    It would be useful here to summarize some basic electricity points which 
    you may know already.

    a) A current flows along a metal or wire when a battery is 
    connected to it.

    b) The current is due to free electrons moving along the metal.

    c) The battery has a potential difference p.d or voltage 
    between its poles due to chemical changes inside the 

    battery. The p.d. pushes the electrons along the metal.

    One pole of the battery is called the positive (+) pole, the other is called the 
    negative (-) pole. The “conventional” current, shown by an arrow, flows 
    in a circuit connected from the + to the – pole. The electrons carrying the 
    current along the circuit wires actually move in the opposite direction to the 
    conventional current but this need not to be taken in account in calculations 
    or circuit formulae.
    C

    Any path along which electrons can flow is a circuit. For a continuous flow 
    of electrons, there must be a complete circuit with no gaps. A gap is usually 
    provided by an electric switch that can be opened or closed to either cut off or 

    allow energy flow.

    Most circuits contain more than one device that receives electric energy 
    from the circuit. These devices are commonly connected in a circuit in one 
    of two ways, series or parallel. When connected in series, the devices and 
    wires connecting them form a single pathway for electron flow between the 
    terminals of the battery, generator or wall socket. When connected in parallel, 
    the devices and wires connecting them form branches, each of which is a 

    separate path for the flow of electrons. 

    Making a series and parallel circuit

    Activity 2 

    Making a series circuit

    Materials:
    * 1 Battery .
    * 3 Bulbs.
    * 3 bulb holders .
    * Assembled battery holder.

    * 4 Pieces of copper wire (as needed).

    Procedure
    1. Construct a complete circuit with a battery and a bulb.
    2. Using another wire, add a second bulb as shown on the picture 

    below.

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    3. What did you notice happened to the first bulb when the second 
    bulb was added?
    _____________________________________________________
    _____________________________________________________
    _____________________________________________________

    4. Look carefully at how the series circuit is set up. Write a prediction 
    of what you think will happen if you unscrew one of the bulbs.
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    Why did you make this prediction? 
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    5. Unscrew bulb “X”. Describe what happens to bulb “Y”.
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    6. Tighten bulb “X”, and unscrew bulb “Y”. Describe what happens 
    to bulb “X”.
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    7. Add a third bulb to your series circuit. What happens to the brightness 
    of the bulbs each time another bulb is added to the series? 
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    8. Add a third bulb to your series circuit. What happens to the brightness 
    of the bulbs each time another bulb is added to the series? 
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    9. Draw a schematic diagram of the circuit you constructed with 

    three bulbs.

    Activity 3

    Making a parallel circuit

    Materials:
    * 1 battery 
    * 3 bulbs
    * Assembled battery holder 3 bulb holders 

    * 6 pieces of copper wire

    Procedure
    1. Construct a complete circuit with one battery and one bulb. 
    2. Using another two wires, add a second bulb as shown in the figure 

    below.

    C

    3. What do you notice happened to the first bulb when the second 
    bulb was added?
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    4. Look carefully at how a parallel circuit is set up. Write a 
    prediction of what you think will happen if you unscrew one of 
    the bulbs in the parallel circuit.
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    Why did you make this prediction? 
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    5. Unscrew bulb “X”. Describe what happens to bulb “Y”.
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    6. Tighten bulb “X” and unscrew bulb “Y”. Describe what happens 
    to bulb “X”.
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    After carrying out experiments for series and parallel circuits, 
    * What advantages and disadvantages can you note for the two cases?
    * What are the characteristics of a series connection and a parallel 

    connection?

    Application activity 5.1
    1. If a battery provides a high voltage, it can ____.
    a. do a lot of work over the course of its lifetime 
    b. do a lot of work on each charge it encounters
    c. push a lot of charge through a circuit
    d. last a long time
    2. Which of the following is true about the electrical circuit in your 
    flashlight?
    a) Charge moves around the circuit very fast - nearly as fast as 
    the speed of light. 
    b) The battery supplies the charge (electrons) which move 
    through the wires.
    c) The battery supplies the charge (protons) which move through 
    the wires.
    d) The charge becomes used up as it passes through the light 
    bulb.
    e) The battery supplies energy which raises charge from low to 
    high voltage.

    f) Nonsense! None of these are true.

    series connections of resistors

    For a series combination of two resistors, the currents are the same in both 
    resistors because the amount of charge that passes through R1 must also pass 

    through R2 and R3 in the same interval of time.

    The potential difference at the generators is equal to the sum of potential 
    difference at each resistor mean that The potential difference applied across 

    the series combination of resistors will divide between the resistors.

    C

    C

    Parallel connection of resistors

    The total current equal to the sum of the current pass through the separate 

    branches.

    The potential difference is the same at each resistor and at the generator or 
    cell.

    C
    C

    Conclusion

    Series and parallel connections each have their own distinctive characteristics.

    In a series circuit, the current is the same at all points; it is not used up. In a parallel 

    circuit the total current equals the sum of the currents in the separate branches.

    C

    Application activity 5.2

    1. As the number of resistors in a series circuit increases, the overall 
    resistance ______ (increases, decreases, remains the same) 
    and the current in the circuit _______ (increases, decreases, 
    remains the same).+
    2. Three resistors are connected in series. If placed in a circuit with 
    a 12 V power supply. Determine the equivalent resistance, the 
    total circuit current, and the voltage drop across and current at 
    each resistor.

    C

    3. Determine the values of the current at and electric potential difference 

    across each of the resistors in a parallel circuit.

    C

    Generators and receptors 

    Electromotive force

    Activity 4

    Electromotive force of a generator

    Materials
    * Battery of 6V,
    * Rheostat
    * Voltmeter
    * Ammeter

    * Connecting wires

    Procedure

    1. Make the connection as shown in the following figure.

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    2. Write down the voltage and the current indicated by the voltmeter 

    and the ammeter when the switch is open.

    3. Close the switch and vary the current in the circuit by varying the 
    values of the rheostat and every time write down values of voltage 

    and current indicated respectively by voltmeter and Ammeter.

    4. Fill in the following table the obtained data:

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    5. When you vary the value of the resistance of the rheostat, does the 

    intensity of current remain constant? Why?

    6. Does the voltage remain constant?

    7. What is the maximum voltage that you have got? How is this 

    voltage called?

    8. In general, if a charge Q (in coulombs) passes through a source of 
    emf E (in volts) which relation will give the electrical energy W

    supplied by the source (in joules)?

    9. Which relation will give the total power of the source?

    Interpretation

    A voltmeter connected to terminals of a battery measures the voltage between 
    terminals of battery. When the switch was closed, we have noticed that there 

    was a current across the circuit and the value of the voltage has been changed.

    By varying the value of the resistance of the rheostat, current in the circuit is 
    changed; voltage indicated by the voltmeter changes also, it decreases when 
    current increases. Its maximum value is reached when the switch is open. 

    Such voltage is called electromotive force E (emf) of the battery.

    The electromotive force emf E of a source (a battery, generator, etc) is the 
    energy transferred to electrical energy when unit charge passes through it. 
    In other words, we can say that the emf of a source of electrical energy is its 

    terminal p.d. on open circuit. 

    The emf of a battery is the maximum possible voltage that the battery can 

    provide between its terminals.

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    Internal resistance

    Activity 5 

    Existence of internal resistance in a generator
    * Consider a certain number of cells which you put in an electric 

    apparatus, like a radio…

    * With your cheek, feel their temperatures before use.

    * Put the cells in your apparatus and let them work for a certain time.

    * Remove the cells and again with your cheek, feel the new 

    temperature of the cells then answer the following questions:

    • Are the two temperatures of the cells equal? (Before and after 
    use)
    • If not, what do you think is the cause of different temperatures?
    • Is one part of the current produced by the generator consumed by 

    it? Why? 

    * The same observations can be made by feeling the temperature of 
    the battery of a telephone before a call and after a call of about 10 
    minutes. Have you felt the increasing of temperature of a phone after 

    using it? If yes, you think it’s due to what? 

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    The term internal resistance refers to the resistance within an emf. The terminal 

    p.d. of a cell on closed circuit is also the p.d. applied to the external circuit.

    In an external circuit electrical energy is changed onto other forms of energy 
    and we regard the terminal p.d. of a cell on closed circuit as being the number 

    of joules of electrical energy changed by each coulomb in the external circuit.

    Not all the electrical energy supplied by a cell to each coulomb is changed 
    in the external circuit. The “lost” energy per coulomb is due to the cell itself 
    having resistance. Each coulomb has to “waste” some energy to get through 
    the cell itself and so less is available for the external circuit. The resistance of 
    a cell is called its internal resistance [r] and depends among other things on 

    its size.

    The electric power dissipated as heat in a cell is given by: Pi = I2 r

    Examples 

    1. The power dissipated as heat in a cell is of 7W, find its internal 
    resistance if a current of 2A flows through it.
    2. Find the power dissipated as heat in a generator of internal 

    resistance 0.6Ω crossed by a current of 3A. 

    Solution

    C

    Remark: Any electrical generator, then, has two important properties, 
    an emf E and an internal resistance r. E and r may be 
    represented separately in a diagram, though in practice they 
    are together between the terminals. To represent a cell, we 
    can write (E, r). So we can think of the battery as an “electric 
    pump”, with its emf E pushing the current round the circuit 
    through both the external (outside) resistor R and internal 

    resistance r

    In an electric circuit, a generator is then represented by the following symbol:

    C

    Activity 6 

    Experiment to find the emf (E) and the internal resistance (r) of a 

    cell

    Materials
    * 1.5V (approx) cell, 
    * Resistance box,
    * Push switch, 

    * Digital Ammeter (0-1A).

    C

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    Relationship between the P.d and the emf at terminals of 

    a cell of a closed circuit

    Activity 7

    Relation between the Emf and the P.d 

    Materials 
    * Dry cell, 
    * Analogy multimetres (2), 

    * Rheostat and switch.

    Procedure

    1. Set up the circuit as shown. 

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    2. Set the resistance of the rheostat to a large value to protect the 
    circuit before switch on the circuit. 
    3. Set the milliammeter to the range 0-1 A or suitable range. 
    4. Set the voltmeter to the range 0-5V or suitable range. 
    5. Switch on the circuit. Record down the readings of the Ammeter 
    and voltmeter. Slide the rider of the rheostat to another position. 
    Record the readings of the ammeter and voltmeter again. Tabulate 

    the results.

    C

    C

    Examples

    1. What is the voltage at terminals of a battery of emf 3V and internal 
    resistance 0.3 Ω when sending a current of 1.5 A in a circuit?

    2. Knowing that the voltage at terminals of a cell is 1.5V and the 
    current crossing the circuit is 1.2 A. Find its emf if its internal 

    resistance is of 0.4 Ω. 

    Solution

    C

    Efficiency of a cell

    Activity 8 

    C


    Since the relation above involves the total power in the circuit, which is the 
    sum of the power supplied in the external circuit and the power dissipated in 
    the cell itself, this one is useful. It’s used to find the efficiency of the cell and 

    how to calculate the voltage at terminals of a cell or battery.

    To calculate the efficiency of the cell.

    C

    Ohm’s law for a circuit having a cell and a 

    resistor

    Activity 9

    From the observation of the following diagram and analysing 

    different elements deduce a relation 

    Questions:

    a) State the Ohm’s law.

    b) Observe the following diagram and list constituting elements.

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    c) Do these elements make an electric circuit? If yes, why?
    d) Are these elements connected in series or in parallel? Why?
    e) How can you find the total resistance of a series connection?
    f) Having the total power supplied by a cell, the power supplied by a 
    cell in an external circuit and the power distributed by the internal 
    resistance. Write the relation between them.
    g) Write down relations for each type of power and substitute them 
    in the relation above. (Powers dissipated in internal and external 
    resistances must be written in terms of resistance)
    h) From the relation found, deduce the emf E. The relation found 
    expresses the Ohm’s law for a circuit having a cell and a resistor.

    i) Express the intensity of the current for this specific case. 

    Pouillet’s law
    We can obtain a form of Ohm’s law by considering a segment of straight 
    wire of uniform cross-sectional area A and length , as shown in Figure 5.8. 
    A potential difference is maintained across the wire, creating in the wire an 

    electric field and a current. 

    C

    The equation representing the dependency of the resistance (R) of a 
    cylindrically shaped conductor (e.g., a wire) upon the variables which affect 

    it, is called Pouillet’s law,

    C


    where 

    • L represents the length of the wire (in m). The longer the wire, the more 
    resistance that there will be

    • A represents the cross-sectional area of the wire (in m2). The wider the wire, 
    the less resistance that there will be to the flow of electric charge. When all 
    other variables are the same, charge will flow at higher rates through wider 
    wires with greater cross-sectional areas than through thinner wires.

    •represents the resistivity of the material (in ). Some materials are better 
    conductors than others and offer less resistance to the flow of charge; they 
    are better conductors. Silver is one of the best conductors but is never used 
    in wires of household circuits due to its cost. Copper and aluminum are 
    among the least expensive materials with suitable conducting ability to 
    permit their use in wires of household circuits. 

    Resistivity of a material is numerically the resistance of a sample of unit 
    length and unit cross-section area, at a certain temperature. The resistivity 
    of a material is dependent upon the material's electronic structure and its 
    temperature. For most (but not all) materials, resistivity increases with 

    increasing temperature. 

    C

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    Application activity 5.3

    1. The total power of a battery is of 9V and its internal resistance 3Ω. 
    Knowing that the current crossed is of 0.4A. Find the efficiency of 
    the battery.

    2. A generator of internal resistance 2Ω sends a current of 4A in a 
    resistor of resistance 10Ω. Calculate its power.

    3. An external resistance of 4Ω is connected to an electric cell of 
    emf 1.5V and internal resistance 2Ω. Calculate the intensity of the 
    current flowing the external resistance.

    4. An electric cell of emf 1.5V and internal resistance 2Ω is 
    connected in series with a resistance of 28Ω. Calculate the power 
    dissipated as heat in the cell. 

    Combination of cells

    Activity 10

    Cells wired in parallel and in series

    Materials:
    * 3 batteries 
    * 2 bulbs 
    * 3 assembled battery holders 
    * 2 bulb holders 
    * 6 pieces of copper wire

    Procedure

    1. Construct a complete circuit with one battery and one bulb.
    2. Observe the brightness of the bulb. 
    3. Construct the circuit below. Are these batteries in series or parallel?
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    How can you tell? 
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

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    4. Observe the brightness of this bulb. Is the bulb brighter than it was 
    with one battery?
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    5. If you added a third battery to this circuit in series, what do you 
    think would happen to the brightness of the bulb? 
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    Why do you think this?
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    6. Add a third battery to this circuit. Describe what happens to the bulb 
    as this battery is added to this circuit in series and why you think the 
    bulb is acting in this way.
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    7. Construct another complete circuit with one battery and one 
    bulb. Record again what the brightness of the bulb is using your 

    brightness metre. 

    8. Look at the pictures below, are the batteries in the picture in series 
    or parallel?
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    How can you tell?
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

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    Construct the circuit in 8. Is the bulb brighter with two batteries 
    than it was with one battery?
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    9. Add one more battery to this circuit in parallel. Describe what 
    happens to the bulb as one more battery is added to this circuit in 
    parallel and why you think the bulb is acting this way.
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    10. Connect then two batteries in opposition to mean the positive 
    (negative) terminals of batteries are connected together and the 
    two free negative (positive) terminals are connected to the bulb. 
    What happens to the bulb?
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    11. Connect two batteries in series in opposition with one battery. 
    When the two free ends of the combination are connected to 
    terminals of the bulb, what happens to the brightness of the bulb?
    _____________________________________________________
    _____________________________________________________

    _____________________________________________________

    Interpretation

    Combination in series

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    When two or more cells are arranged in series, the total emf is the algebraic 
    sum of their emfs and the total internal resistance is the algebraic sum of their 

    internal resistances

    C

    Note: A series arrangement is used to increase the voltage, also the 
                total internal resistance of the circuit, so the energy loss due to 

                internal resistance is greater than for a single cell.

    C

    Examples

    1. Four 1.5V cells are connected in series to a 12Ω lightbulb. If the 
    resulting current is 0.45A, what is the internal resistance of each 

    cell, assuming they are identical and neglecting wires?

    2. A certain number of cells of emf 1.5V and internal resistance 
    2Ω are connected in series. When connected this combination to 
    an external resistance of 10Ω, a current of 500mA flows in this 

    resistance. Find the number of cells used. 

    C

    Combination in opposition

    Let us see also what the result could be if the cells were associated in opposition 

    C

    Let us combine two cells in opposition. Two terminals of same sign are 
    connected together. The direction of the current in the circuit will be determined 
    by the direction of the current produced by the cell having the higher emf. For 

    internal resistances they are in series. So we write: 

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    Note: You might think that connecting batteries in opposition would 
    be wasteful. For more purposes, that will be true. But such 
    an opposition arrangement is precisely how a battery charger 

    works.

    Example

    A cell of emf 2V and internal resistance 0.2Ω is associated in opposition 
    with another cell of emf 1.5V and internal resistance 1.2Ω. Calculate the 

    intensity of the current knowing that the external resistance is 1.1Ω. 

    Solution

    C

    Combination in parallel

    C

    C

    Note: The parallel arrangement is useful normally only if the emfs are 
    the same. A parallel arrangement is not used to increase the 
    voltage, but rather to provide large currents. Each of the cells 
    in parallel has to produce only a fraction of the total current, 
    so the energy loss due to internal resistance is less than for a 

    single cell; the batteries will be exhausted less quickly.

    Examples

    1. We have 8 cells of emf 1.5V and internal resistance 2Ω. Calculate 
    the intensity of the current flowing in an external resistance of 1Ω 

    connected to the terminals of the 8 cells combined in parallel.

    2. Six cells of unknown emf and internal resistance of 2Ω are 
    associated in parallel. When an external resistance of 1Ω is 
    connected to this combination a current of 1.5A is produced. 

    Calculate the emf.

    Solution

    C

    Mixing series and parallel combinations

    C

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    Example

    Four cells of emf 4.5V each and internal resistance 2Ω are combined in 
    series. The combination is connected to an external resistance of 24Ω
    a) What is the intensity of the current?
    b) Same question if the cells are combined in parallel.
    c) Same question if the combination has two parallel series of two 

    cells each. 

    Solution

    C

    Receptors

    Activity 11

    Distinguishing a receptor from a passive resistor

    a) Observe the following devices and name them

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    b) What is the use of each one?
    c) The flowing of the current in them produces the same 
    effect? Explain.
    d) Among them, which ones transform the whole electric 
    energy consumed in heat and which ones transform a part of 
    electric energy consumed in another kind of energy which is 
    not heat? 
    e) As we had in the case of generators, what are characteristics 

    of these apparatuses?

    Conclusion: Among the apparatuses above, there are some which transform 
    the total electric energy consumed into heat and some transform just a part 
    into heat, other part transformed into another type of energy which is not heat. 
    Those which transform the whole quantity of electric energy consumed into 
    heat are passive resistors or passive receptors and those transforming a part of 
    the consumed electric energy in another form of energy which is not heat are 

    called receptors or active receptors. 

    The main characteristics are back electromotive force and internal resistance. 

    Back electromotive force

    Back electromotive force (emf) is normally used to refer to the voltage that 

    is developed in electric motors. This is due to the relative motion between the 

    magnetic field from the motor's field windings and the armature of the motor!

    Internal resistance

    The internal resistance of a receptor r’ is its ability to oppose electric current. 
    When a receptor is traversed by an electric current, part of the energy consumed 
    is transformed into heat. The power dissipated in the receptor by joule effect 
    is: PJ = I 2r

    The p.d at terminals of a receptor

    Activity 12

    Find the P.d at terminals of a motor

    Materials
    * Electric motor
    * Ammeter
    * Voltmeter

    * Power supply

    Procedure

    1. Make the connection as shown in the figure below.

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    Measure the voltage (V) between terminals of the motor (M) and 

    the current I in the circuit. 

    Questions
    a) What is the net electrical power received by the motor?
    b) What becomes this power and how is it transformed?
    c) What is the relation between the voltage and the back 
    electromotive force?
    d) From the relation found, how do you calculate the intensity 

    of the current flowing?

    Application activity 5.4

    1. A circuit has in series a generator of emf 6V and internal 
    resistance 0.1Ω, a receptor of back emf 1.5V and internal 
    resistance 0.4Ω and a passive resistor of 8.5Ω. Calculate:
    a) The intensity of the current flowing in the circuit.
    b) The power supply by the generator.

    c) The quantity of heat produced in the resistor in one minute.

    2. A battery has an emf of 12.0V and an internal resistance of 0.05Ω. 
    Its terminals are connected to a load resistance of 3.00Ω. (a) Find 
    the current in the circuit and the terminal voltage of the battery. 
    (b) Calculate the power delivered to the load resistor, the power 
    delivered to the internal resistance of the battery, and the power 

    delivered by the battery.

    3. Calculate the terminal voltage for a battery with an internal 
    resistance of 0.9Ω and an emf of 8.5V when the battery is connected 

    in series with (a) an 81Ω resistor, and (b) 810Ω.

    4. A 9V battery whose internal resistance r is 0.5Ω is connected in 

    the circuit shown in the figure. 

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    a) How much current is drawn from the source?
    b) What is the terminal voltage of the battery? 

    c) What is the current in the 6Ω resistor?

    5. What is the internal resistance of a 12V car battery whose terminal 
    voltage drops to 8.4V when the starter draws 75A? What is the 

    resistance of the starter?

    6. A 1.5V dry cell can be tested by connecting it to a low-resistance 
    Ammeter. It should be able to supply atleast 22A. What is the 
    internal resistance of the cell in this case, assuming it is much 

    greater than that of the Ammeter?

    7. A cell whose terminals are connected to a wire in nickel silver of 
    resistivity 30 x 10 -6 Ωcm and cross sectional area 0.25mm2
     and 
    length 5m sends a current of 160mA. When the length is reduced 
    to a half, the intensity of the current is of 300mA. Calculate:
    a) The internal resistance.

    b) The emf of the cell.

    8. A cell (E = 1.5V, r = 1.3Ω) sends a current in an external 
    resistance of 3Ω. Calculate: 
    a) The intensity of the current in the circuit.
    b) The p.d at terminals of the cell.
    c) The power of generator.

    d) The efficiency of the cell.

    9. A battery is composed by 120 cells in series. Each element has an 
    emf of 2V and an internal resistance of 0.001Ω. The combination 
    is connected to an external resistance of 4.8Ω. Calculate: 
    a) The intensity of the current in the circuit.
    b) The voltage at terminals of the battery.
    c) The energy dissipated by joule effect when the current 

    flows in the circuit in one hour.

    Kirchhoff’s rules 

    Activity 13

    Find the equivalent Resistance

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    In this experiment you will be using a digital multimeter (DMM) which 
    can function as either a voltmeter or an Ammeter. 
    The voltmeter must always be wired in parallel with the resistor whose 
    voltage you are measuring. The ammeter, used to measure current, must 
    always be wired in series. Disconnect the meter from the circuit before 
    you change the function setting. Failure to follow these procedures can 
    result in serious damage to the meter. Be sure that you use the correct units 

    with your data.

    Materials
    * 1 multimeter.
    * 1 330 Ω or 240 Ω resistor.
    * 1 1000 Ω (1KΩ) resistor.
    * 1 2000 Ω (2KΩ) resistor.
    * 1 3000 Ω (3KΩ) resistor or 1 3300 Ω (3.3KΩ) resistor. 
    * 1 0- 10K resistor substitution box.
    * 2 spade lugs.
    * 2 2’ red banana wires.
    * 2 2’ black banana wires. 

    * 4 4” black banana wires. 

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    Simple circuits can be analysed using the expression V = IR and the rules 
    for series and parallel combinations of resistors. Very often, however, it is 
    not possible to reduce a circuit to a single loop. The procedure for analysing 
    more complex circuits is greatly simplified if we use two principles called 

    Kirchhoff's rules:

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    Kirchhoff’s second rule follows from the law of conservation of energy. Let us 
    imagine moving a charge around a closed loop of a circuit. When the charge 
    returns to the starting point, the charge–circuit system must have the same 
    total energy as it had before the charge was moved. The sum of the increases in 
    energy as the charge passes through some circuit elements must equal the sum 
    of the decreases in energy as it passes through other elements. The potential 
    energy decreases whenever the charge moves through a potential drop -IR 
    across a resistor or whenever it moves in the reverse direction through a source 
    of emf. The potential energy increases whenever the charge passes through a 

    battery from the negative terminal to the positive terminal.

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    Examples

    1. A single-loop circuit contains two resistors and two batteries, as 
    shown in figure 5.29 (neglect the internal resistances of the batteries). 
    (a) Find the current in the circuit. (b) What power is delivered to 

    each resistor? What power is delivered by the 12V battery?

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    We therefore need only two loop equations to determine the unknown 

    currents. (The third loop equation would give no new information).

    Applying Kirchhoff’s loop rule to loops ABCDA and BEFCB and 

    traversing these loops clockwise, we obtain the expressions

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    END UNIT ASSESSMENT

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    13. A dead battery is charged by connecting it to the live battery of 
    another car with jumper cables as shown in the figure. Determine 

    the current in the starter and in the dead battery.

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    ◄Unit 4 Work, Energy and PowerUnit 6 : Sources of Energy in the World►