Question

A 50 V d.c power supply is used to charge a battery of eight lead accumulators, each of emf 2 V and internal resistance $1/8\Omega$. The charging current also runs a motor connected in series with the battery. The resistance of the motor is $5\Omega$ and the steady current supply is 4 A. The total power lost due to heat dissipation is

• A. 64 W
• B. 80 W
• C. 90 W
• D. 96 W

Answer 1

Answer: (D)

96 W

Total power supplied $=50V\times 4A=200W$

Power lost as heat to the battery $=I^{2}R$$=4^2\times (8\times \frac{1}{8})=16W$

Power lost as heat to the motor $=I^{2}r$$=4^2\times 5=80W$

Hence, total power lost as heat=$96W$