Question 180 of 380

Two cells are connected between points $A$ and $B$ as shown. Cell 1 has emf of $12V$ and internal resistance of $3\Omega$. Cell 2 has emf of $6V$ and internal resistance of $6\Omega$. An external resistor $R$ of $4\Omega$ is connected across $A$ and $B$. The current flowing through $R$ will be ______ $A$.

Your answerCheck answer

Show answer and solution▾

Answer: 1

Equivalent internal resistance of two cells in parallel is ${r}_{eq}=\dfrac{{r}_{1}{r}_{2}}{{r}_{1}+{r}_{2}}=\dfrac{3\times 6}{3+6}=2\Omega$
Equivalent emf of given cells
${E}_{eq}=\left(\dfrac{{E}_{1}}{{r}_{1}}-\dfrac{{E}_{2}}{{r}_{2}}\right){r}_{eq}=\left(\dfrac{12}{3}-\dfrac{6}{6}\right)2=6V$
Given: $R=4\Omega$
The circuit can be redrawn as shown below.

Current through resistor $R$ is $I=\dfrac{6}{2+4}=1A$.