Question

Three resistances of equal values are arranged in four different configurations as shown below. Power dissipation in the increasing order is:

• A. (III)$<$(II)$<$(IV)$<$(I)
• B. (II)$<$(III)$<$(IV)$<$(I)
• C. (I)$<$(IV)$<$(III)$<$(II)
• D. (I)$<$(III)$<$(II)$<$(IV)

Answer 1

The correct option is A (III)$<$(II)$<$(IV)$<$(I)

Since the current in all the configurations is given to be same, power dissipation is given by

$P=I^2{R}_{eq.}$

$R_{Ieq.}=R+R+R=3R$

$P_I=3I^{2}R$

$R_{IIeq.}=(R+R)||R=\frac{\left(2R\right)\left(R\right)}{2R+R}=\frac{2R}{3}$

$P_{II}=\frac{2}{3}{I}^{2}R$

$R_{IIIeq.}=R||R||R=\frac{R}{3}$

$P_{III}=\frac{1}{3}{I}^{2}R$

$R_{IVeq.}=\left(R||R\right)+R=\frac{R}{2}+R=\frac{3R}{2}$

$P_{IV}=\frac{3}{2}{I}^{2}R$

Thus, correct sequence is $III<II<IV<I$