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Question

(C) To find the equivalent resistance between points $A$ and $D$,we analyze the circuit step-by-step.$1$. The circuit consists of resistors each of $1

Vedclass · Accepted Answer

$30$

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(C) To find the equivalent resistance between points $A$ and $D$,we analyze the circuit step-by-step.$1$. The circuit consists of resistors each of $10 \ \Omega$.$2$. Looking at the right side,the resistors $R_3$ and $R_4$ are in series,giving $10 \ \Omega + 10 \ \Omega = 20 \ \Omega$.$3$. Similarly,the top branch resistors $R_2$ and the $10 \ \Omega$ resistor connected to $C$ are in series,giving $10 \ \Omega + 10 \ \Omega = 20 \ \Omega$.$4$. These two $20 \ \Omega$ branches are in parallel with each other,resulting in an equivalent resistance of $\frac{20 	imes 20}{20 + 20} = 10 \ \Omega$.$5$. Finally,this $10 \ \Omega$ equivalent resistance is in series with the remaining $10 \ \Omega$ resistors $R_1$ and $R_6$ (or the path leading to $A$ and $D$).$6$. Following the simplification shown in the image,the total resistance between $A$ and $D$ is $10 \ \Omega + 10 \ \Omega + 10 \ \Omega = 30 \ \Omega$.

The equivalent resistance between the points $A$ and $D$ is ................ $\Omega$.

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