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  • In the circuit shown in figure. is a variable <img alt="" src="https://cdn.entrance360.com/me

In the circuit shown in figure. (A), R_3 is a variable


resistance. As the value R_3 is changed, current  I  though the cell varies as shown. Obviously, the variation is asymptotic, i.e. I \rightarrow 6 \mathrm{~A} as R_3 \rightarrow \infty. Resistances R_1 and R_2 are, respectively -
 

Option: 1

4 \Omega, 2 \Omega


Option: 2

2 \Omega, 4 \Omega


Option: 3

2 \Omega, 2 \Omega


Option: 4

1 \Omega, 4 \Omega


Answers (1)

best_answer

For R_3=0, I=9 A (from the graph). In this situation, the circuit can be drawn as shown in figure (A).

Here I=\frac{36}{R_1+R_2}=9
or\mathrm{R}_1+\mathrm{R}_2=4 \quad \quad \quad \dots(1)

For \mathrm{R}_3 \rightarrow \infty, equivalent resistance of \mathrm{R}_2 \text{ and }\mathrm{R}_3 in parallel will be
\begin{aligned} \frac{1}{R_{e q}} & =\frac{1}{R_2}+\frac{1}{R_3 \rightarrow \infty} \\ & =\frac{1}{R_2} \end{aligned}
[Current through R_2 will be zero because we have a short circuit \left(R_3=0\right. ) across \mathrm{R}_2]

or \mathrm{R}_{\mathrm{eq}}=\mathrm{R}_2
In this situation, the circuit can be drawn as shown figure (B).


\begin{aligned} \text { Now } I & =\frac{36}{R_1+2 R_2} \\ & =6 \quad \quad (given) \\ \therefore R_1 & +2 R_2=6 \quad \quad \dots(2) \end{aligned}
from eqs. (1) and (2)
\mathrm{R}_1=2 \Omega, \mathrm{R}_2=2 \Omega

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seema garhwal

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