Two cells are connected between points A and B as shown. Cell 1 has an emf of 12 V and internal resistance of 3. Cell 2

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Two cells are connected between points A and B as shown. Cell 1 has an emf of 12 V and internal resistance of 3. Cell 2 has an emf of 6 V and internal resistance of 6. An external resistor of 4 is…

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Given: Two cells are connected between A and B. Their emfs and internal resistances are $12 \, \text{V}, 3\Omega$ and $6 \, \text{V}, 6\Omega$ respectively. The external resistor is $R = 4\Omega$ .

Find: The current through $R$ .

The parallel cell combination is first replaced by its equivalent source.

E_{\text{eq}} = \frac{\frac{12}{3} - \frac{6}{6}}{\frac{1}{3} + \frac{1}{6}}

E_{\text{eq}} = 6 \, \text{V}

r_{\text{eq}} = 2\Omega

Now the circuit becomes a $6 \, \text{V}$ source with internal resistance $2\Omega$ in series with the external resistor $4\Omega$ .

Using Ohm's law,

i = \frac{6}{2+4}

i = 1 \, \text{A}

Therefore, the current through the resistor is $1 \, \text{A}$ .

Equivalent source derivation image showing the original parallel cell network between A and B and expressions for equivalent emf, equivalent resistance, and resistor value.

Simplified equivalent circuit between A and B with a 6 V source, 2 ohm internal resistance, 4 ohm external resistor, and current i marked in the loop.

Given: The two cells are connected in parallel opposition across A and B with internal resistances $3\Omega$ and $6\Omega$ .

Find: Current in the external resistor $4\Omega$ .

For parallel sources with internal resistances, compute the equivalent emf and equivalent internal resistance first.

E_{\text{eq}} = \frac{\frac{E_1}{r_1} - \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{12}{3} - \frac{6}{6}}{\frac{1}{3} + \frac{1}{6}}

E_{\text{eq}} = \frac{4 - 1}{\frac{1}{2}} = 6 \, \text{V}

Also,

r_{\text{eq}} = \frac{r_1 r_2}{r_1 + r_2} = \frac{3 \times 6}{3+6} = 2\Omega

This equivalent source supplies the resistor $R = 4\Omega$ .

Total resistance in the loop is

R_{\text{total}} = 2 + 4 = 6\Omega

Hence,

i = \frac{E_{\text{eq}}}{R_{\text{total}}} = \frac{6}{6} = 1 \, \text{A}

Therefore, the required current is $1 \, \text{A}$ .

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