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Question: A network of four \(20\mu F\) capacitors is connected to a 600 V supply as shown in the figure. The ...

A network of four 20μF20\mu F capacitors is connected to a 600 V supply as shown in the figure. The charge on capacitors C1C_{1} and C4C_{4} are

A

(a) 4×10−3C,12×10−3C4 \times 10^{- 3}C,12 \times 10^{- 3}C

A

(b) 6×10−3C,12×10−3C6 \times 10^{- 3}C,12 \times 10^{- 3}C

A

(c) 2×10−3C,4×10−3C2 \times 10^{- 3}C,4 \times 10^{- 3}C

A

(d) 3×10−3C,2×10−3C3 \times 10^{- 3}C,2 \times 10^{- 3}C

Explanation

Solution

(a): Charge on each capacitor, C1,C2,C3C_{1,}C_{2,}C_{3}is same = q let charge on C4C_{4}be q′q'

∴qC1+qC2+qC3=600\therefore\frac{q}{C_{1}} + \frac{q}{C_{2}} + \frac{q}{C_{3}} = 600

Or q[1C1+1C2+1C3]=600q\left\lbrack \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} \right\rbrack = 600

Or qCs=600\frac{q}{C_{s}} = 600 (using (i))

∴q=600Cs=600×203μC=4000×10−6C=4×10−3C\therefore q = 600C_{s} = 600 \times \frac{20}{3}\mu C = 4000 \times 10^{- 6}C = 4 \times 10^{- 3}CAlso q′C4=600\frac{q'}{C_{4}} = 600

Or q′=C4×600μC=20×600μC=12×10−3Cq' = C_{4} \times 600\mu C = 20 \times 600\mu C = 12 \times 10^{- 3}C