A battery of $6 \; V$ and internal resistance $0.5\; \Omega$ is joined in parallel with another of $10\;V$ and internal resistance $1\;\Omega$. The combination sends a current through an external resistance of $12\; \Omega$. Find the current through each battery.

Given,

Emf of one cell $(E_1) = 6\; V$

Emf of another cell $(E_2) = 10\; V$

Internal resistance of one cell $(r_1) = 0.5 \; \Omega$

Internal resistance of another cell $(r_2) = 1 \; \Omega$

External resistance $(R) = 12 \; \Omega$

Current through $E_1 \; (I_1) = ?$

Current through $E_2 \; (I_2) = ?$

Using Kirchhoff's voltage law in closed-loop ABXCDA, we have

$E_1 = I_1\; r_1 + I_3 \; R$

$6 = I_1\;r_1 + (I_1 + I_2)\;R$

or, $6 = 0.5\;I_1 + 12\;I_1  + 12 \; I_2$

or, $6 = 12.5 \; I_1 + 12 \; I_2$ .......... (i)

Again, Using Kirchhoff's voltage law in closed loop ABYCDA, we have

$E_2 = I_2\; r_2 + I_3 \; R$

or, $10 = I_2 \; r_2 + (I_1 + I_2)\;R$

or, $10 = 12 \; I_1 + 13\; I_2$ .......... (ii)

By Solving equation (i) and (ii) we get

$I_1 = -2.27 \; A$       and       $I_2 = 2.86 \; A$

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