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You have been provided with a resistance box, a voltmeter, a key, a battery, and other necessary materials.

  • Connect the circuit as shown in the diagram above.
  • With the key K closed, read and record the voltmeter reading V\(_{0}\).
  • Set the resistance R in the resistance box equal to I\(\Omega\).
  •  Close the key, read and record the potential difference V on the voltmeter.
  • EvaluateR\(^{-1}\) and V\(^{-1}\) .
  • Repeat the procedure for five other values or R = 2\(\Omega\), 3\(\Omega\), 4\(\Omega\), 5\(\Omega\) and 6\(\Omega\).
  • Tabulate your readings.
  • Plot a graph with V\(^{-1}\) on the vertical axis and R on the horizontal axis.
  • Determine the slope, s, of the graph and the intercept, C, on the vertical axis.
  • Evaluate c\(^{-1}\)
  • State two precautions taken to obtain accurate results.

(b)i. Define the potential difference between two points in an electric circuit.

ii. Explain why the emf of a cell is greater than the p.d. across the call when it is supplying Current through an external resistance.

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6Tam
1 year ago

The e.m.f. of a cell is the p.d. across the terminals when it is in an open circuit i.e. hot supplying current to an external circuit. While terminal p.d. 5 is defined as the p.d, between the terminals of a cell when it is delivering current to an external circuit. Therefore, emf is greater because it has not yet been connected to an external circuit.

This doesn't explain why the emf of a cell is greater than the p.d. across the call when it is supplying Current through an external resistance. 🤦

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