Three capacitors 3 μ F , 6 μ F and 6 μ F are connected in series to a source of 120 V. The potential difference, in volt, across the 3 μ F capacitor will be

Question

Three capacitors 3 μ F , 6 μ F and 6 μ F are connected in series to a source of 120 V. The potential difference, in volt, across the 3 μ F capacitor will be (A) 24 (B) 30 (C) 40 (D) 60

Tardigrade Admin · Accepted Answer

The combination of three charges in series

\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}

\frac{1}{C} = \frac{1}{3} + \frac{1}{6} + \frac{1}{6}

\Rightarrow C = \frac{6}{4} = 1.5 μ F

The charge of this circuit

q = C V = 1.5 \times 120

q = 180 μ C

The potential difference across the

3 μ F

q = C V

V = \frac{q}{C} = \frac{180}{3} = 60 V

Q. Three capacitors 3μF,6μF and 6μF are connected in series to a source of 120V. The potential difference, in volt, across the 3μF capacitor will be

24

30

40

60

The combination of three charges in series C1​=C1​1​+C2​1​+C3​1​ C1​=31​+61​+61​ ⇒C=46​=1.5μF The charge of this circuit q=CV=1.5×120 q=180μC The potential difference across the 3μF q=CV V=Cq​=3180​=60V

Q. Three capacitors $3 μ F, 6 μ F$ and $6 μ F$ are connected in series to a source of $120 V$ . The potential difference, in volt, across the $3 μ F$ capacitor will be