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You are provided a cell of emf, E, a key, K, an ammeter, A, a voltmeter, V, a standard resistor, R, a jockey, J, a potentiometer PF and connecting wires. Use the diagram below as a guide to carry out the following instructions.

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3(a)

  • (i) Determine the approximate value of E using the voltmeter. Connect the voltmeter directly across the terminals of the cell E with the key K open. The reading on the voltmeter will give the approximate electromotive force (EMF), E, of the cell.

  • (ii) - (viii) Procedure for experiment. These are instructions for performing the experiment and recording data.

  • (ix) Evaluate Q=V1V2Q = \frac{V_1}{V_2}, W=IuW = Iu and W−1W^{-1}. These are calculations to be performed for each set of readings.

  • (x) Repeat the experiment for u=40,60,80 and 100 cmu = 40, 60, 80 \text{ and } 100 \text{ cm}. For each case, record V1,V2,Q,I,W and W−1V_1, V_2, Q, I, W \text{ and } W^{-1}. This is an instruction for data collection.

  • (xi) Tabulate the results. The table should have the following columns:

    uu (cm)V1V_1 (V)V2V_2 (V)II (A)Q=V1V2Q = \frac{V_1}{V_2}W=IuW = Iu (A cm)W−1W^{-1} (A−1^{-1} cm−1^{-1})
    20
    40
    60
    80
    100

  • (xii) Plot a graph of W−1W^{-1} on the vertical axis and QQ on the horizontal axis. This is an instruction for you to perform.

  • (xiii) Calculate the slope, ss, of the graph. This is an instruction for you to perform.

  • (xiv) Given that σ=Es\sigma = Es, calculate the value σ\sigma.

  1. The quantities QQ and W−1W^{-1} are defined as Q=V1V2Q = \frac{V_1}{V_2} and W−1=1IuW^{-1} = \frac{1}{Iu}.

  2. From the circuit, V1=IRV_1 = IR and V2=IρuV_2 = I \rho u, where ρ\rho is the resistance per unit length of the potentiometer wire.

  3. Substitute these into the expression for QQ

    $$
Q = \frac{IR}{I \rho u} = \frac{R}{\rho u}
$$

  4. Rearrange to find uu

    $$
u = \frac{R}{\rho Q}
$$

  5. Substitute this expression for uu into W−1W^{-1}

    $$
W^{-1} = \frac{1}{I \left(\frac{R}{\rho Q}\right)} = \frac{\rho Q}{IR}
$$

  6. This shows a linear relationship W−1=(ρIR)QW^{-1} = \left(\frac{\rho}{IR}\right) Q.

    The slope ss of the graph of W−1W^{-1} vs QQ is: s=ρIR s = \frac{\rho}{IR}

  7. Substitute ss into the given formula for σ\sigma

    $$
\sigma = Es = E \left(\frac{\rho}{IR}\right)
$$
The unit of $Q$ is dimensionless. The unit of $W^{-1}$ is $\text{A}^{-1}\text{cm}^{-1}$. Therefore, the unit of the slope $s$ is $\text{A}^{-1}\text{cm}^{-1}$.  
The unit of $\sigma = Es$ is $\text{V} \cdot \text{A}^{-1}\text{cm}^{-1} = \frac{\text{V}}{\text{A}} \cdot \text{cm}^{-1} = \Omega \cdot \text{cm}^{-1}$.  
The unit of $\sigma$ is $\boxed{\Omega \text{ cm}^{-1}}$.

    • (xv) Use the graph to determine W when Q = 5. Hence, determine I if u = 50 cm.

      1. Locate Q=5Q=5 on the horizontal axis of your graph.
      2. Read the corresponding value of W−1W^{-1} from the vertical axis.
      3. Calculate W=1W−1W = \frac{1}{W^{-1}}.
      4. Using the formula W=IuW = Iu, and given u=50 cmu = 50 \text{ cm}, calculate I=W50 cmI = \frac{W}{50 \text{ cm}}.

    • (xvi) State two precautions taken to ensure accurate results when performing this experiment.

      1. Ensure clean terminals and tight connections throughout the circuit to minimize contact resistance.
      2. Avoid prolonged current flow by opening the key K between readings to prevent heating of the potentiometer wire and the cell, which could alter their resistance and EMF respectively.

    3(b)

    • (i) State two reasons why Manganin is used for making potentiometer wire.

      1. Manganin has a very low temperature coefficient of resistance, meaning its resistance changes minimally with temperature fluctuations, ensuring a stable potential gradient.
      2. Manganin has a high resistivity, allowing a long wire with significant resistance to be made compact, which is ideal for a potentiometer.

    • (ii) The current drawn by an electric device is 1.2 A when it is connected to a potential difference of 240.0 V. Calculate the current drawn by the same device if it is connected to a 110.0 V source.

  8. Calculate the resistance of the device using Ohm's Law.

    $$
R = \frac{V_1}{I_1}
$$
$$
R = \frac{240.0 \text{ V}}{1.2 \text{ A}}
$$
$$
R = 200 \text{ } \Omega
$$

  9. Calculate the current drawn when connected to a 110.0 V source.

    $$
I_2 = \frac{V_2}{R}
$$
$$
I_2 = \frac{110.0 \text{ V}}{200 \text{ } \Omega}
$$
$$
I_2 = 0.55 \text{ A}
$$
The current drawn by the device is $\boxed{0.55 \text{ A}}$.
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