In the circuit,R_{1} = R_{2} = R. The value o | vedclass

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(A) Let $R_{1} = R_{2} = R$. The current through $R_{2}$ is $(I - 1.5) \ A$. Applying Kirchhoff's voltage law $(KVL)$ in the loop containing $E, R_{1}

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$180 \ V, 60 \ \Omega$

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(A) Let $R_{1} = R_{2} = R$. The current through $R_{2}$ is $(I - 1.5) \ A$. Applying Kirchhoff's voltage law $(KVL)$ in the loop containing $E, R_{1}$,and $R_{2}$:$E - I R_{1} - (I - 1.5) R_{2} = 0$Since $R_{1} = R_{2} = R$,we have $E = I R + (I - 1.5) R = R(2I - 1.5) \quad ... (i)$Applying $KVL$ in the outer loop containing $E, R_{1}$,and $R'$:$E - I R_{1} - 1.5 R' = 0$$E = I R + 1.5 R' \quad ... (ii)$From the provided circuit diagram,the voltage across $R_{2}$ is the same as the voltage across $R'$,so $V_{R_{2}} = V_{R'}$.$(I - 1.5) R = 1.5 R'$$R' = \frac{(I - 1.5) R}{1.5}$Substituting $R'$ into equation $(ii)$:$E = I R + 1.5 \left[ \frac{(I - 1.5) R}{1.5} ight] = I R + (I - 1.5) R = R(2I - 1.5)$This confirms the consistency. Given the options,let's test $R = 60 \ \Omega$ and $E = 180 \ V$:$180 = 60(2I - 1.5) \Rightarrow 3 = 2I - 1.5 \Rightarrow 2I = 4.5 \Rightarrow I = 2.25 \ A$.Then $I - 1.5 = 2.25 - 1.5 = 0.75 \ A$.Voltage across $R_{2} = 0.75 	imes 60 = 45 \ V$.Voltage across $R' = 1.5 	imes R' = 45 \ V \Rightarrow R' = 30 \ \Omega$.This is a valid physical circuit. Thus,$E = 180 \ V$ and $R_{1} = 60 \ \Omega$ is the correct pair.

In the circuit,$R_{1} = R_{2} = R$. The value of $E$ and $R_{1}$ are $\ldots \ldots \ldots$ ($E$ = $EMF$,$R_{1}$ = resistance).

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