Question

Three resistor are connected in parallel across 100-volt supply. One resistor absorb 1600 watts and the second 1200 watts. The total energy used by the whole circuit in 10 minutes is 276x10^4 joules. Find resistance of the third resistor.

Answer 1

Given:

$V=100\:\rm V$

$P_1=1600\:\rm W$

$P_2=1200\:\rm W$

$t=\rm 10\: min=600\: s$

$W=276*10^4\: \rm J$

The net power absorbed from supply

$$P=P_1+P_2+P_3=W/t$$

The power absorbed by third resistor

$$P_3=W/t-P_1+P_2=V^2/R_3$$

Hence, the resistance of third resistor

$$R_3=\frac{V^2}{W/t-P_1-P_2}$$

$$R_3=\frac{100^2}{276*10^4/600-1600-1200}=5.6\:\rm \Omega$$