A physics class investigates the terminal p.d. of a cell by connecting it to a resistor R with an ammeter and a voltmeter as shown in Fig. 1.1. (a) The e.m....
Question 1 Report
A physics class investigates the terminal p.d. of a cell by connecting it to a resistor R with an ammeter and a voltmeter as shown in Fig. 1.1.
(a) The e.m.f. of the cell marked on its casing is 1.5 V. Record this value. [1]
(b) Record the ammeter and voltmeter readings. The ammeter reads 0.30 A and the voltmeter reads 1.38 V. [1]
(c) State whether the voltmeter reading equals the e.m.f. of the cell. [1]
(d) Explain why the terminal p.d. is less than the e.m.f. [1]
(e) Calculate the resistance of R using the ammeter and voltmeter readings. [1]
(f) Calculate the internal resistance of the cell. Show your working. [2]
Answer Details
(a) e.m.f. of the cell: 1.5 V. [1]
(b) Ammeter: 0.30 A; Voltmeter: 1.38 V. [1]
(c) No. The voltmeter reading (1.38 V) is less than the e.m.f. (1.5 V). [1]
(d) The cell has internal resistance \( r \). When current flows, some of the e.m.f. is used to drive current through this internal resistance. [1]
The "lost volts" equal \( V_{\text{lost}} = Ir \). The terminal p.d. is therefore \( V = \varepsilon - Ir \), which is always less than the e.m.f. when current is drawn.
(e) Resistance of R:
\[ R = \frac{V}{I} = \frac{1.38}{0.30} = 4.6 \; \Omega \] [1]
(f) Lost volts:
\[ V_{\text{lost}} = \varepsilon - V = 1.5 - 1.38 = 0.12 \text{ V} \] [1]
Internal resistance:
\[ r = \frac{V_{\text{lost}}}{I} = \frac{0.12}{0.30} = 0.40 \; \Omega \] [1]
The internal resistance is small compared to R (0.40 Ω vs 4.6 Ω), which is why most of the e.m.f. appears across the external resistor.
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