Two capacitors of 3 μ F and 6 μ F are connected in series across a potential difference of 120 V. Then the potential difference across 3 μ F capacitor is

Question

Two capacitors of 3 μ F and 6 μ F are connected in series across a potential difference of 120 V. Then the potential difference across 3 μ F capacitor is (A) 50 V (B) 60 V (C) 70 V (D) 80 V

Tardigrade Admin · Accepted Answer

Here, (1/CS)=(1/3)+(1/6)=(1/2) CS=2 μ F Q=CSV=2×120 =240 μ C V1=(Q/C1)=(240/3) =80 V

Q. Two capacitors of $3 μ F$ and $6 μ F$ are connected in series across a potential difference of $120 V$ . Then the potential difference across $3 μ F$ capacitor is

$50 V$

$60 V$

$70 V$

$80 V$

Here, $\frac{1}{C _{S}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$
$C_{S} = 2 μ F$
$Q = C_{S} V = 2 \times 120$
$= 240 μ C$
$V_{1} = \frac{Q}{C _{1}} = \frac{240}{3}$
$= 80 V$

Q. Two capacitors of 3μF and 6μF are connected in series across a potential difference of 120V. Then the potential difference across 3μF capacitor is

50V

60V

70V

80V