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A supply of 400V is connected across capacitors of 3μf and 6μf in series. Cal...

Question

A supply of 400V is connected across capacitors of 3μf and 6μf in series. Calculate the charge

Options

A) 8 x 10$$^{-4}$$C

B) 4 x 10$$^{-2}$$C

C) 8 x 10$$^{-3}$$C

D) 4 x 10$$^{-8}$$C

Explanation:

 C$$_T$$ = C$$_1$$ × C$$_2$$ C$$_1$$ + C$$_2$$
 = 3 × 6 3 + 6

= $$\frac{18}{9}$$ = 2μf
Q = CV
⇒ 2 × 10$$^{-6}$$ × 400
⇒ 800 × 10$$^{-6}$$C = 8 × 10$$^{-4}$$C

Dicussion (1)

•  C$$_T$$ = C$$_1$$ × C$$_2$$ C$$_1$$ + C$$_2$$
 = 3 × 6 3 + 6

= $$\frac{18}{9}$$ = 2μf
Q = CV
⇒ 2 × 10$$^{-6}$$ × 400
⇒ 800 × 10$$^{-6}$$C = 8 × 10$$^{-4}$$C