SSC JE Electrical Engineering 22 Jan 2018 Morning Paper

Question : 107 of 200

Marks: +1, -0

Determine the value of change stored (in mC) in a capacitor, when the value of capacitance is 0.01 mF and the potential difference between the ends of the capacitor is 20 V.

Validate

Solution:

Given data
C = 0.01 mf
V = 20 V
Now charge stored in capacitor is given by
Q = CV = (0.01 × 20)V

=(1/100×20)V

= 0.2 V

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