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The diagram illustrates an electrical circuit used to investigate a science experiment. The rheostat is adjusted to vary the current and voltage in the circuit. Fig 2 (a) and Fig 2 (b) below illustrate ammeter readings I=I1, I2, I3, I4, I5 and voltmeter readings v=v1, v2, v3, v4, and v5, in table 1 below respectively when the key is closed.

Physics

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Step-by-step solution

4 steps
  1. Read and record the ammeter readings from Fig.

    2(a).

    The meter scales show the following values for the needle positions:

    ILIMIR0.20 A0.40 A0.60 A\begin{array}{|c|c|c|} \hline I_\mathrm{L} & I_\mathrm{M} & I_\mathrm{R} \\ \hline 0.20~\mathrm{A} & 0.40~\mathrm{A} & 0.60~\mathrm{A} \\ \hline \end{array}
  2. Read and record the voltmeter readings from Fig.

    2(b).

    The meter scales show the following values for the needle positions (in volts):

    VLVMVR1.0 V2.0 V3.0 V\begin{array}{|c|c|c|} \hline V_\mathrm{L} & V_\mathrm{M} & V_\mathrm{R} \\ \hline 1.0~\mathrm{V} & 2.0~\mathrm{V} & 3.0~\mathrm{V} \\ \hline \end{array}
  3. Verify Ohm's law using V=IRV = IR.

    For low setting:
    R=VLIL=1.00.20=5.0 ΩR = \frac{V_\mathrm{L}}{I_\mathrm{L}} = \frac{1.0}{0.20} = 5.0~\Omega

    For medium setting:
    R=VMIM=2.00.40=5.0 ΩR = \frac{V_\mathrm{M}}{I_\mathrm{M}} = \frac{2.0}{0.40} = 5.0~\Omega

    For high setting:
    R=VRIR=3.00.60=5.0 ΩR = \frac{V_\mathrm{R}}{I_\mathrm{R}} = \frac{3.0}{0.60} = 5.0~\Omega

    Average resistance R=5.0 ΩR = 5.0~\Omega.

  4. Plot the graph with II (vertical axis, in A) vs VV (horizontal axis, in V).

    Plot the points: (1.0,0.20)(1.0, 0.20), (2.0,0.40)(2.0, 0.40), (3.0,0.60)(3.0, 0.60).

    The graph is a straight line through origin with slope 1R=0.20 A/V\frac{1}{R} = 0.20~\mathrm{A/V}.

    R = 5.0 \Omega

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