A battery is connected to a variable resistor. When the current is 0.40 A, the terminal p.d. is 5.6 V. When the current is 0.75 A, the terminal p.d. is 4.9 V.
(a) Calculate the internal resistance and the emf of the battery. [3]
(b) In a required practical, a student records several pairs of values of terminal p.d. $V$ and current $I$. Describe how she should use a graph to find the emf and internal resistance. [2]
- (a) $\varepsilon = V + Ir$: $\varepsilon = 5.6 + 0.40r$ and $\varepsilon = 4.9 + 0.75r$ allow $r = \frac{\Delta V}{\Delta I} = \frac{5.6-4.9}{0.75-0.40}$(1)
- (a) $0.35r = 0.70$ so $r = 2.0$ Ω(1)
- (a) $\varepsilon = 5.6 + 0.40 \times 2.0 = 6.4$ V allow 4.9 + 0.75 × 2.0; ecf for their r(1)
- (b) plot $V$ (y-axis) against $I$ (x-axis); emf = the y-intercept allow $I$ against $V$ with emf = intercept on the $V$ axis(1)
- (b) internal resistance = −gradient (magnitude of gradient) allow for an $I$ against $V$ graph: $r$ = −1/gradient(1)
- Answer: (a) $r = 2.0$ Ω, $\varepsilon = 6.4$ V (b) $\varepsilon$ = intercept on V axis; $r$ = −gradient
- Examiner tip: Write ε = V + Ir for both readings and solve simultaneously — the emf is the same in both.