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(C) The circuit consists of four $4 \Omega$ resistors forming a square. When a battery is connected across diagonally opposite corners,the circuit act

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(C) The circuit consists of four $4 \Omega$ resistors forming a square. When a battery is connected across diagonally opposite corners,the circuit acts as two parallel branches,each containing two $4 \Omega$ resistors in series.Resistance of each branch = $4 \Omega + 4 \Omega = 8 \Omega$.Since there are two such branches in parallel,the equivalent external resistance $R_{	ext{ext}}$ is:$R_{	ext{ext}} = \frac{8 \Omega 	imes 8 \Omega}{8 \Omega + 8 \Omega} = 4 \Omega$.The total resistance of the circuit including the internal resistance $r = 2 \Omega$ is:$R_{	ext{total}} = R_{	ext{ext}} + r = 4 \Omega + 2 \Omega = 6 \Omega$.The total power dissipated in the circuit is given by $P = \frac{E^2}{R_{	ext{total}}}$,where $E = 12 	ext{ V}$.$P = \frac{(12)^2}{6} = \frac{144}{6} = 24 	ext{ W}$.

Four $4 \Omega$ resistors are connected together along the edges of a square. $A$ $12 \text{ V}$ battery with an internal resistance of $2 \Omega$ is connected across a pair of the diagonally opposite corners of the square. The power dissipated in the circuit is (in $\text{ W}$)

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