In the circuit shown,the variable resistance | vedclass

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Question

(C) Let the main battery voltage be $V$ and the variable resistance be $R_v$.When the key is at position $1$,the circuit consists of the main battery

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$10$

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(C) Let the main battery voltage be $V$ and the variable resistance be $R_v$.When the key is at position $1$,the circuit consists of the main battery and the variable resistance $R_v$ in series with the ideal ammeter. The current $I = 2 \, A$ is given by $I = \frac{V}{R_v} = 2 \, A$,so $V = 2 R_v$.When the key is at position $2$,the circuit includes the battery $E$ and resistance $x$ in series with the main battery and $R_v$. Since the ammeter reading remains $2 \, A$,the net $EMF$ in the circuit is $V - E$ (assuming they are in opposition) or $V + E$. Given the circuit diagram,the battery $E$ opposes the main battery. Thus,$I = \frac{V - E}{R_v} = 2 \, A$.Substituting $V = 2 R_v$,we get $\frac{2 R_v - E}{R_v} = 2$,which implies $2 - \frac{E}{R_v} = 2$,meaning $\frac{E}{R_v} = 0$,which is impossible unless $R_v$ is infinite. However,looking at the circuit,when the key is at position $2$,the current $I = 2 \, A$ flows through $x$ and $E$. The voltage drop across the branch containing $E$ and $x$ must be equal to the voltage drop when the key is at position $1$. Actually,the condition implies that the potential difference across the terminals $1$ and $2$ is $E + Ix = 0$ is not correct. The correct interpretation is that the current $I$ is constant. In position $1$,$V = I R_v$. In position $2$,$V - E = I(R_v + x)$. Since $I$ is the same,$V = I R_v$ and $V - E = I R_v + I x$. Substituting $V = I R_v$ into the second equation: $I R_v - E = I R_v + I x$,which gives $-E = I x$. Taking magnitudes,$E = I x$. Thus,$x = \frac{E}{I} = \frac{20 \, V}{2 \, A} = 10 \, \Omega$.

In the circuit shown,the variable resistance is adjusted such that the ammeter reading is the same in both positions $1$ and $2$ of the key. The reading of the ammeter is $2 \, A$. If $E = 20 \, V$,then $x$ is: ................... $\Omega$

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