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(A) For circuit $(a)$,the $100 \, \Omega$ and $200 \, \Omega$ resistors are in series,and this combination is in series with the parallel combination

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(A) For circuit $(a)$,the $100 \, \Omega$ and $200 \, \Omega$ resistors are in series,and this combination is in series with the parallel combination of $300 \, \Omega$ and $R$. However,looking at the diagram $(a)$,the $100 \, \Omega$ and $200 \, \Omega$ are in series,and the $300 \, \Omega$ and $R$ are in parallel with each other. The equivalent resistance is $R_{eq} = (100 + 200) + \frac{300R}{300+R} = 300 + \frac{300R}{300+R} = \frac{90000 + 600R}{300+R}$.The total current is $I = \frac{10}{R_{eq}} = \frac{10(300+R)}{90000 + 600R}$.The voltage across $R$ is $V_a = I 	imes \frac{300R}{300+R} = \frac{10(300+R)}{90000 + 600R} 	imes \frac{300R}{300+R} = \frac{3000R}{600R + 90000} = \frac{5R}{R + 150}$.For circuit $(b)$,the $200 \, \Omega$ and $R$ are in series,and this is in parallel with $300 \, \Omega$. The equivalent resistance of this parallel part is $R_p = \frac{300(200+R)}{300+200+R} = \frac{60000 + 300R}{500+R}$.The total resistance is $R_{eq} = 100 + R_p = 100 + \frac{60000 + 300R}{500+R} = \frac{50000 + 100R + 60000 + 300R}{500+R} = \frac{110000 + 400R}{500+R}$.The total current is $I = \frac{10}{R_{eq}} = \frac{10(500+R)}{110000 + 400R}$.The voltage across the parallel combination is $V_p = I 	imes R_p = \frac{10(500+R)}{110000 + 400R} 	imes \frac{60000 + 300R}{500+R} = \frac{600000 + 3000R}{110000 + 400R} = \frac{6000 + 30R}{1100 + 4R}$.The voltage across $R$ is $V_b = V_p 	imes \frac{R}{200+R} = \frac{30(200+R)}{1100+4R} 	imes \frac{R}{200+R} = \frac{30R}{1100+4R}$.Equating $V_a = V_b$: $\frac{5R}{R+150} = \frac{30R}{1100+4R} \Rightarrow \frac{1}{R+150} = \frac{6}{1100+4R} \Rightarrow 1100 + 4R = 6R + 900 \Rightarrow 2R = 200 \Rightarrow R = 100 \, \Omega$.

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