Question

Two batteries of emf 3V 6V and internal resistances 0.2Ω 0.4Ω are connected in parallel. This combination is connected to a 4Ω resistor. Find:
(i) the equivalent emf of the combination
(ii) the equivalent internal resistance of the combination
(iii) the current drawn from the combination

Hint:

In parallel circuits, the equivalent emf is a weighted average, and the equivalent resistance is the reciprocal of the sum of reciprocals of individual resistances.

Answer 1

(i) The equivalent emf \( E_{\text{eq}} \) of two batteries connected in parallel is given by: \[ E_{\text{eq}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \] where \( E_1 = 3V \), \( E_2 = 6V \), \( r_1 = 0.2 \, \Omega \), and \( r_2 = 0.4 \, \Omega \). Substituting the values: \[ E_{\text{eq}} = \frac{(3)(0.4) + (6)(0.2)}{0.2 + 0.4} = \frac{1.2 + 1.2}{0.6} = \frac{2.4}{0.6} = 4V \] (ii) The equivalent internal resistance \( r_{\text{eq}} \) of two batteries in parallel is given by: \[ r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2} \] Substituting the values: \[ r_{\text{eq}} = \frac{(0.2)(0.4)}{0.2 + 0.4} = \frac{0.08}{0.6} = 0.1333 \, \Omega \] (iii) The total resistance in the circuit is \( R_{\text{total}} = r_{\text{eq}} + 4 \, \Omega = 0.1333 + 4 = 4.1333 \, \Omega \). The current drawn from the combination is given by Ohm's law: \[ I = \frac{E_{\text{eq}}}{R_{\text{total}}} = \frac{4}{4.1333} = 0.968 \, \text{A} \] Thus, the current drawn from the combination is approximately \( 0.968 \, \text{A} \).