Question

In the circuit shown in the figure (A), $R_{3}$ is a variable resistance. As the value $R_{3}$ is changed, the current $I$ through the cell varies as shown in the figure (B). As $R_{3} \rightarrow \infty$ , the current $I \, \rightarrow \, 6 \, A$ . What is the value of the sum of the resistances $R_{1}$ and $R_{2}$ in ohms?

Answer 1

For $R_{3}=0,I=9 \, A$ (from the graph).In this situation, the circuit can be drawn as shown in figure $\left(A\right)$ .

[current through R₂ will be zero because we have a short circuit $\left(R_{3} = 0\right)$ across $R_{2}$ ]
Here $I=\frac{36}{R_{1} + R_{2}}=9 \, A$
$R_{1}+R_{2}=4 \, \Omega $