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 mechanical power stored in the motor is

• A. 80 W
• B. 40 W
• C. 64 W
• D. 30 W

Answer 1

Answer: (B)

40 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

$=200W-96W=104W$

Chemical power stored in battery $=$ Voltage $\times$ current

$=16V\times 4A=64W$

Hence, mechanical power available $=104W-64W=40W$