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 useful power available is

• A. 96 W
• B. 200 W
• C. 104 W
• D. 100 W

Answer 1

Answer: (C)

104 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 $=96W$

Hence the usable power $=$ Total power supplied-Power lost as heat

$=200W-96W=104W$