A capacitor with a capacitance of 50µf when connected to a battery of 400 V. The charge and energy stored on it is? a. 0.05 C and 5 J b. 0.05 C and 10 J c. 0.02 C and 4 J d. 0.08 C and 12 J

c. 0.02 C and 4 J

Explanation:

Applying,

Q = CV Equation 1

Where Q = Charge, C = Capacitance of the capacitor, V = Voltage.

From the question,

Given: C = 50 μF = 50×10⁻⁶ F, V = 400 V

Substitute these values into equation 1

Q = (50×10⁻⁶)(400)

Q = 0.02 C.

Also Applying

E = CV²/2 Equation 2

Where E = Energy stored.

Therefore,

E = (50×10⁻⁶ )(400²)/2

E = 4 J

Hence the right option is c. 0.02 C and 4 J

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need to see circuit to get time constant