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End Of Chapter Exercises

Exercise 10.3

[IEB 2001/11 HG1] - Emf

  1. Explain the meaning of each of these two statements:

    1. “The current through the battery is \(\text{50}\) \(\text{mA}\).”

    2. “The emf of the battery is \(\text{6}\) \(\text{V}\).”

  2. A battery tester measures the current supplied when the battery is connected to a resistor of \(\text{100}\) \(\text{Ω}\). If the current is less than \(\text{50}\) \(\text{mA}\), the battery is “flat” (it needs to be replaced). Calculate the maximum internal resistance of a \(\text{6}\) \(\text{V}\) battery that will pass the test.

\begin{align*} V & = IR \\ \mathcal{E} - Ir = IR \\ r = \frac{IR - \mathcal{E}}{-I} \\ = \frac{(\text{5,0} \times \text{10}^{-\text{2}})(\text{100}) - (6)}{-(\text{5,0} \times \text{10}^{-\text{2}})} \\ = \frac{-1}{-0.05} \\ = \text{20}\text{ Ω} \end{align*} \(\text{20}\) \(\text{Ω}\)

[IEB 2005/11 HG] The electric circuit of a torch consists of a cell, a switch and a small light bulb, as shown in the diagram below.

f3f2c5cab3666d0bb0fec5d0667ccdcc.png

The electric torch is designed to use a D-type cell, but the only cell that is available for use is an AA-type cell. The specifications of these two types of cells are shown in the table below:

Cell

emf

Appliance for which it is designed

Current drawn from cell when connected to the appliance for which it is designed

D

\(\text{1,5}\) \(\text{V}\)

torch

\(\text{300}\) \(\text{mA}\)

AA

\(\text{1,5}\) \(\text{V}\)

TV remote control

\(\text{30}\) \(\text{mA}\)

What is likely to happen and why does it happen when the AA-type cell replaces the D-type cell in the electric torch circuit?

What happens

Why it happens

(a)

the bulb is dimmer

the AA-type cell has greater internal resistance

(b)

the bulb is dimmer

the AA-type cell has less internal resistance

(c)

the brightness of the bulb is the same

the AA-type cell has the same internal resistance

(d)

the bulb is brighter

the AA-type cell has less internal resistance

The internal resistance of the type AA cell is greater than that for the type D cell. The bulb will be dimmer.

The correct answer is option A.

[IEB 2005/11 HG1] A battery of emf ε and internal resistance r = \(\text{25}\) \(\text{Ω}\) is connected to this arrangement of resistors.

97a27cb9d41c277ac8c19a1d9e37bbbb.png

The resistances of voltmeters \({V}_{\text{1}}\) and \({V}_{\text{2}}\) are so high that they do not affect the current in the circuit.

  1. Explain what is meant by “the emf of a battery”.

    The power dissipated in the \(\text{100}\) \(\text{Ω}\) resistor is \(\text{0,81}\) \(\text{W}\).

  2. Calculate the current in the \(\text{100}\) \(\text{Ω}\) resistor.

  3. Calculate the reading on voltmeter \({V}_{\text{2}}\).

  4. Calculate the reading on voltmeter \({V}_{\text{1}}\).

  5. Calculate the emf of the battery.

Solution not yet available

[SC 2003/11] A kettle is marked \(\text{240}\) \(\text{V}\); \(\text{1 500}\) \(\text{W}\).

  1. Calculate the resistance of the kettle when operating according to the above specifications.

  2. If the kettle takes \(\text{3}\) \(\text{minutes}\) to boil some water, calculate the amount of electrical energy transferred to the kettle.

  1. Explain what is meant by “the emf of a battery”:

    The emf of a battery is the maximum electrical potential energy it can impart to a unit charge in a circuit.

    The power dissipated in the \(\text{100}\) \(\text{Ω}\) resistor is \(\text{0,81}\) \(\text{W}\).

  2. Calculate the current in the \(\text{100}\) \(\text{Ω}\) resistor:

    We can use the power dissipated in a known resistor to determine the current through the resistor:

    \begin{align*} P&= I^2R \\ (\text{0,81})&= I^2(100) \\ I&=\sqrt{\frac{\text{0,81}}{100}} \\ &=\text{0,09}\text{ A} \end{align*}
  3. Calculate the reading on voltmeter \({V}_{\text{2}}\).

    \({V}_{\text{2}}\) is the potential difference across the parallel combination. We know the current so we can determine the resistance and then the potential difference. The equivalent resistance is:

    \begin{align*} \frac{1}{R_P}&= \frac{1}{R_1} + \frac{1}{R_2} \\ \frac{1}{R_P}&= \frac{1}{50} + \frac{1}{50} \\ \frac{1}{R_P}&= \frac{2}{50} \\ R_p&= \text{25}\text{ Ω} \end{align*}

    The total current through the equivalent resistance is the same as we calculated as the parallel network is in series with the \(\text{100}\) \(\text{Ω}\) resistor. Therefore the potential difference across the parallel combination is:

    \begin{align*} V_P&= IR_P \\ &= \text{0,09}\text{25} \\ &= \text{2,25}\text{ V} \end{align*}
  4. Calculate the reading on voltmeter \({V}_{\text{1}}\).

    We can calculate the potential difference across the \(\text{100}\) \(\text{Ω}\) and add it to the potential difference across the parallel combination to determine the potential difference measured by \({V}_{\text{1}}\).

    \begin{align*} V_{100}&= IR \\ &= \text{0,09}(100) \\ &= \text{9}\text{ V} \end{align*} \begin{align*} {V}_{\text{1}}&= V_{100} + V_P \\ &= \text{9}+\text{2,25} \\ & \text{11,25}\text{ V} \end{align*}
  5. Calculate the emf of the battery.

    The emf of the battery will be \({V}_{\text{1}}\), the potential difference across the load added to the potential difference across the internal resistance.

    \begin{align*} V_r&= Ir \\ &= \text{0,09}\text{25} \\ &= \text{2,25}\text{ V} \end{align*} \begin{align*} \mathcal{E}&= \text{2,25} + \text{11,25} \\ &= \text{13,50}\text{ V} \end{align*}
\(\text{0,09}\) \(\text{A}\); \(\text{2,25}\) \(\text{V}\); \(\text{11,25}\) \(\text{V}\) ; \(\text{13,50}\) \(\text{V}\)

[IEB 2001/11 HG1] - Electric Eels

Electric eels have a series of cells from head to tail. When the cells are activated by a nerve impulse, a potential difference is created from head to tail. A healthy electric eel can produce a potential difference of \(\text{600}\) \(\text{V}\).

  1. What is meant by “a potential difference of \(\text{600}\) \(\text{V}\)”?

  2. How much energy is transferred when an electron is moved through a potential difference of \(\text{600}\) \(\text{V}\)?

The work done on a charge \(q\) when moving through a potential difference \(V\) is:

\begin{align*} W&=Vq \\ &= (600)(\text{1,6} \times \text{10}^{-\text{19}} \\ &= (600)(\text{1,6} \times \text{10}^{-\text{19}} \\ &= \text{9,6} \times \text{10}^{-\text{17}}\text{ J} \end{align*} \(\text{9,6} \times \text{10}^{-\text{17}}\) \(\text{J}\)

The diagram shows an electric circuit consisting of a battery and four resistors.

The potential difference (voltage) over the battery is \(V_A = \text{7,6}\text{ V}\)

The the resistors are rated as follows:

  • \(R_1 = \text{4,7}\text{ Ω}\)
  • \(R_2 = \text{6,9}\text{ Ω}\)
  • \(R_3 = \text{4,9}\text{ Ω}\)
  • \(R_4 = \text{4,3}\text{ Ω}\)

Assume that positive charge is flowing in the circuit (conventional current).

Using the concepts of Ohm's law, and electric circuits, determine the following:

What type of circuit is shown in the diagram?

We need to determine whether the circuit is a series, parallel, or combination type of circuit. We do this by looking at how current flows through a circuit.

Recall that current is the movement of electric charge from a higher potential to a lower potential.

We assume that the flowing charge is positive (conventional current). This means that the charges start at the positive terminal of our power source (battery). Here the charges have lots of electrical potential energy.

The charges then move towards the negative terminal through the path that is created by the components and wires of the circuit. During this process, the electrical potential energy is converted into thermal energy by the resistors. Therefore, the charges have less electrical potential energy when they reach the negative terminal of the power source (battery).

The flow of current is indicated by the arrows in the diagram below:

From the diagram it is clear that there is only one path for the current to flow, since the circuit does not split into two or more paths. This means that the current flows through all the components, one after the other (in series). We will label this current \(I_A\).

The circuit shown in the diagram is a series circuit.

\(0\)

What is the total equivalent resistance \(R_{eq}\) of the circuit?

  • round your answer to 1 digit after the decimal comma
  • use the values for any physical constants you might need, as listed here

We know that the circuit in the diagram is a series type circuit.

Recall that in a series circuit the equivalent resistance can be calculated by summing the resistance values of the individual resistors:

\[R_{eq,S} = R_1 + R_2 + R_3 + \ldots ~~~~~\text{(1)}\]

We can rewrite \(\text{(1)}\) for the circuit in question as:

\[R_{eq} = R_1 + R_2 + R_3 + R_4 ~~~~~\text{(2)}\]

We can represent this by drawing an equivalent circuit:

We are given the following information:

  • the resistance of \(R_1 = \text{4,7}\text{ Ω}\)
  • the resistance of \(R_2 = \text{6,9}\text{ Ω}\)
  • the resistance of \(R_3 = \text{4,9}\text{ Ω}\)
  • the resistance of \(R_4 = \text{4,3}\text{ Ω}\)

Substituting the above values in \(\text{(2)}\), we get:

\begin{align*} R_{eq} & = R_1 + R_2 + R_3 + R_4\\ & = \text{4,7} + \text{6,9} + \text{4,9} + \text{4,3}\\ & = \text{20,8}\text{ Ω} \end{align*}

Therefore, the total equivalent resistance in the circuit, or \(R_{eq} = \text{20,8}\text{ Ω}\)

\(20.8\)

Question 3

What is the potential difference (voltage) across \(R_1\), or \(V_1\) ?

  • round your answer to 3 digits after the decimal comma
  • use the values for any physical constants you might need, as listed here

Potential difference, or voltage is a way to describe the difference in electrical potential energy across a component in a circuit. Remember that a resistor converts electrical potential energy into thermal energy, so the electrical potential is higher on the "in" side compared to the "out" side of a resistor.

Ohm's law describes the relationship between the total current \(~I~\) through an ohmic conductor, its resistance \(R\), and the potential difference \(V~\) across it:

\[I = \frac{V}{R}~~~~~\text{(3)}\]

For the equivalent circuit, we can rewrite \(\text{(3)}\) as:

\[I_A = \frac{V_A}{R_{eq}}~~~~~\text{(4)}\]

We are asked to calculate the potential difference \(V_{1}\) over \(R_{1}\), as shown in the circuit diagram above, so we use Ohm's law and rewrite \(\text{(3)}\) to get:

\[V_1 = I_1R_1~~~~~\text{(5)}\]

Since the circuit in question is a series circuit, we know that the current \(I_{1}\) flowing through \(R_{1}\), is the same as the current \(I_{A}\). This is represented as:

\[I_1 = I_A~~~~~\text{(6)}\]

Substituting \(\text{(6)}\) into \(\text{(5)}\) we get:

\[V_1 = I_AR_1~~~~~\text{(7)}\]

Substituting \(\text{(4)}\) into \(\text{(7)}\) we get:

\[V_{1} = I_AR_1 = \left(\frac{V_A}{R_{eq}}\right)R_1 = \frac{V_AR_1}{R_{eq}}~~~~~\text{(8)}\]

Recall from Question 2:

\[R_{eq} = R_1 + R_2 + R_3 + R_4 ~~~~~\text{(2)}\]

Substituting \(\text{(2)}\) into \(\text{(8)}\) we get:

\[V_{1} = \frac{V_AR_1}{R_1 + R_2 + R_3 + R_4}~~~~~\text{(9)}\]

We are given the following information:

  • the resistance of \(R_1 = \text{4,7}\text{ Ω}\)
  • the resistance of \(R_2 = \text{6,9}\text{ Ω}\)
  • the resistance of \(R_3 = \text{4,9}\text{ Ω}\)
  • the resistance of \(R_4 = \text{4,3}\text{ Ω}\)
  • the potential difference (voltage) over the battery \(V_A = \text{7,6}\text{ V}\)

Substituting the above values in \(\text{(9)}\), we get:

\begin{align*} V_{1} & = \frac{V_AR_1}{R_1 + R_2 + R_3 + R_4}\\ & = \frac {(\text{7,6})(\text{4,7})} {\text{4,7} + \text{6,9} + \text{4,9} + \text{4,3}}\\ & = 1,71731 \ldots~~~~~\text{(calculated)}\\ & \approx \text{1,717}\text{ V}~~~~~\text{(rounded)} \end{align*}

Therefore, the potential difference (voltage) across \(R_1\), or \(V_1 \approx \text{1,717}\text{ V}\).

\(1.717\)

What is the potential difference (voltage) across \(R_2\), \(R_3\), and \(R_4\), or \(V_2\), \(V_3\), and \(V_4\)?

  • round your answers to 3 digits after the decimal comma
  • use the values for any physical constants you might need, as listed here

We are asked to calculate the three remaining potential differences, \(V_2\), \(V_3\), and \(V_4\). This is shown in the diagram below:

We will use the same method to obtain three expression for \(V_2\), \(V_3\), and \(V_4\) as was used in Question 3 for \(V_1\). We get:

\[V_{2} = \frac{V_AR_2}{R_1 + R_2 + R_3 + R_4}~~~~~\text{(10)}\]

\[V_{3} = \frac{V_AR_3}{R_1 + R_2 + R_3 + R_4}~~~~~\text{(11)}\]

\[V_{4} = \frac{V_AR_4}{R_1 + R_2 + R_3 + R_4}~~~~~\text{(12)}\]

We are given the following information:

  • the resistance of \(R_1 = \text{4,7}\text{ Ω}\)
  • the resistance of \(R_2 = \text{6,9}\text{ Ω}\)
  • the resistance of \(R_3 = \text{4,9}\text{ Ω}\)
  • the resistance of \(R_4 = \text{4,3}\text{ Ω}\)
  • the potential difference (voltage) over the battery \(V_A = \text{7,6}\text{ V}\)

Substituting the above values in \(\text{(10)}\), \(\text{(11)}\), and \(\text{(12)}\) we get:

\begin{align*} V_{2} & = \frac{V_AR_2}{R_1 + R_2 + R_3 + R_4}\\ & = \frac {(\text{7,6})(\text{6,9})} {\text{4,7} + \text{6,9} + \text{4,9} + \text{4,3}}\\ & = 2,52115 \ldots~~~~~\text{(calculated)}\\ & \approx \text{2,521}\text{ V}~~~~~\text{(rounded)} \end{align*}

\begin{align*} V_{3} & = \frac{V_AR_3}{R_1 + R_2 + R_3 + R_4}\\ & = \frac {(\text{7,6})(\text{4,9})} {\text{4,7} + \text{6,9} + \text{4,9} + \text{4,3}}\\ & = 1,79038 \ldots~~~~~\text{(calculated)}\\ & \approx \text{1,790}\text{ V}~~~~~\text{(rounded)} \end{align*}

\begin{align*} V_{4} & = \frac{V_AR_4}{R_1 + R_2 + R_3 + R_4}\\ & = \frac {(\text{7,6})(\text{4,3})} {\text{4,7} + \text{6,9} + \text{4,9} + \text{4,3}}\\ & = 1,57115 \ldots~~~~~\text{(calculated)}\\ & \approx \text{1,571}\text{ V}~~~~~\text{(rounded)} \end{align*}

Therefore, the potential difference (voltage) across \(R_2\), or \(V_2 \approx \text{2,521}\text{ V}\).

The potential difference (voltage) across \(R_3\), or \(V_3 \approx \text{1,790}\text{ V}\).

The potential difference (voltage) across \(R_4\), or \(V_4 \approx \text{1,571}\text{ V}\).

\(2.521\) \(1.790\) \(1.571\)