Looking for classes? Ksquare Career Institute, Bengaluru →
A wire of resistance $R$ is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:
A
$\frac{R}{64}$
B
$\frac{R}{32}$
C
$\frac{R}{16}$
D
$\frac{R}{8}$
Explanation
Each piece has resistance $R/8$. Four in parallel give $R_p = (R/8)/4 = R/32$. Two such sets in series give $R_{net} = R/32 + R/32 = R/16$.
Detailed Solution
Resistance of each piece: $r = R/8$. When 4 pieces are connected in parallel, the equivalent resistance of one set is $R_p = \frac{r}{4} = \frac{R/8}{4} = \frac{R}{32}$. When two such identical parallel sets are connected in series, the net resistance is $R_{net} = R_p + R_p = \frac{R}{32} + \frac{R}{32} = \frac{2R}{32} = \frac{R}{16}$.