Question

Two batteries of emf's \(3V \& 6V\) and internal resistances 0.2  Ω \(\&\) 0.4  Ω  are connected in parallel. This combination is connected to a 4 Ω 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:

When batteries are connected in parallel, the equivalent emf is a weighted average of the individual emfs, and the equivalent internal resistance is smaller than the smallest individual internal resistance.

Answer 1

For two batteries connected in parallel, the equivalent electromotive force \( E_{\text{eq}} \) and equivalent internal resistance \( r_{\text{eq}} \) are calculated as follows: \[ E_{\text{eq}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} \] \[ r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2} \] Given values:
\( E_1 = 3 \, \text{V} \), \( r_1 = 0.2 \, \Omega \)
\( E_2 = 6 \, \text{V} \), \( r_2 = 0.4 \, \Omega \)

(i) Equivalent emf calculation: \[ E_{\text{eq}} = \frac{(3 \times 0.4) + (6 \times 0.2)}{0.2 + 0.4} = \frac{1.2 + 1.2}{0.6} = \frac{2.4}{0.6} = 4 \, \text{V} \]

(ii) Equivalent internal resistance calculation: \[ r_{\text{eq}} = \frac{0.2 \times 0.4}{0.2 + 0.4} = \frac{0.08}{0.6} = 0.133 \, \Omega \] 

(iii) Current drawn from the combination: The total circuit resistance is: \[ R_{\text{total}} = r_{\text{eq}} + R = 0.133 + 4 = 4.133 \, \Omega \] The current \( I \) is determined by: \[ I = \frac{E_{\text{eq}}}{R_{\text{total}}} = \frac{4}{4.133} = 0.968 \, \text{A} \] ---