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Question

The power dissipated in circuit is given by $P=\frac{V^{2}}{R}$

For the resistances in parallel the equivalent resistance is less than smallest res

Vedclass · Accepted Answer

Both $(A)$ and $(C)$

Vedclass · Answer

The power dissipated in circuit is given by $P=\frac{V^{2}}{R}$

For the resistances in parallel the equivalent resistance is less than smallest resistance in combination of resistors hence power dissipation is maximum in this case.

For the resistances in series the equivalent resistance is addition of resistors hence power dissipation is minimum in this case.

Hence, to avoid maximum power dissipation in parallel circuit R must be equal to zero so the equivalent of parallel resistors will be

$R_{p}=\frac{(0)(8)}{8+0}=0 \Omega$

since, $R_{p}=0 \Omega$ voltage across it is zero.

Therefore power dissipated in $2 \Omega$ resistor is

$P=\frac{V^{2}}{R}$

$P=\frac{(12)^{2}}{2}$

$P=72 W$

The value of the resistance $R$ in figure is adjusted such that power dissipated in the $2\,\Omega$ resistor is maximum. Under this condition

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