Question

A .25uF capicitor is charged by a battery. After being charged, the capacitor is connected to a small electric motor. Assuming 100% effency what initial voltage must the capicitor have if the motor is to lift a 15g mass through a height of .8 cm.

Answer 1

From the given question ,I have given mass, m = 15 g = 0.015 kg

height lifted, h = 0.8 cm = 0.008 m

capacitance of capacitor, C = 0.25 uF 0.25 * 10^-6 uF

Let the capacitor is charged with battery having voltage V volts

then the energy stored in capacitor = 0.5*C*V²

As the efficiency is 100 %

there energy stored in capacitor = energy dissipated in motor during lifting mass,m to the height 0.8 cm

therefore 0.5*0.25*10^-6*V² = m*g*h where g = 10 m/s² is the accelaration due to gravity

0.125*10^-6*V² = 0.015*10*0.008

V² = 1200*10^-6/(0.125*10^-6)

V² = 1200000/125

V² = 9600

V = 97.97 volts

hence the required voltage is V = 97.97 volts

ANSWER : required voltage is V = 97.97 volts