The effective resistance between P and Q for | vedclass

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(B) Looking at the circuit,the two $3 \Omega$ resistors at the top are in series. Their equivalent resistance is $R_1 = 3 \Omega + 3 \Omega = 6 \Omega

Vedclass · Accepted Answer

$12$

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(B) Looking at the circuit,the two $3 \Omega$ resistors at the top are in series. Their equivalent resistance is $R_1 = 3 \Omega + 3 \Omega = 6 \Omega$.This $6 \Omega$ equivalent resistance is in parallel with the $6 \Omega$ resistor connected between the two branches. The equivalent resistance $R_2$ of this parallel combination is given by $\frac{1}{R_2} = \frac{1}{6 \Omega} + \frac{1}{6 \Omega} = \frac{2}{6 \Omega} = \frac{1}{3 \Omega}$,so $R_2 = 3 \Omega$.Now,the circuit consists of the $4 \Omega$ resistor,the $3 \Omega$ equivalent resistance $(R_2)$,and the $5 \Omega$ resistor,all connected in series.Therefore,the total effective resistance between $P$ and $Q$ is $R_{eq} = 4 \Omega + 3 \Omega + 5 \Omega = 12 \Omega$.

The effective resistance between $P$ and $Q$ for the following network is: (in $\Omega$)

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