Question

Q.16 A variable capacitor is connected to a 200V battery. If its capacitance is changed from 2 to X micrfarad, the decrease in energy of the capacitor is 2 x 10^(-2). The value of X is -

A. 1 microfarad
B. 2 microfarad
C. 3 microfarad
D. 4 microfarad

Answer 1

Answer:

A) 1 MICROFARDAD

Explanation:

Answer 2

Given - Voltage: 200 volts

Change in energy: - 2 x 10^(-2)

Original capacitance: 2 microfarad

Find - New capacitance

Solution - Change in capacitance is related with voltage and energy through the formula -

∆U = 1/2V²(C2 - C1)

Taking C2 or new capacitance as X. Keeping the values in the formula to find the value of X.

Multiplying 2 on Left Hand Side

-2*10-²*2 = 200²*(X - 2)

-4*10-²/4*10⁴ = X - 2*10-⁶

-1*10-⁶ = X - 2*10-⁶

X = 2*10-⁶ - 1*10-⁶

X = 1*10-⁶ Farad

This can be further written as X = 1 microfarad. Since 10-⁶ is equal to micro units.

Hence, 1 microfarad is the new capacitance.