Two capacitors of capacity 6 μ F and 12 μ F in series are connected by potential of 150 V . The potential of the capacitor of capacity 12 μ F will be

Question

Two capacitors of capacity 6 μ F and 12 μ F in series are connected by potential of 150 V . The potential of the capacitor of capacity 12 μ F will be (A) 25V (B) 50V (C) 100 V (D) 150 V

Tardigrade Admin · Accepted Answer

Given, C1=6μ F,C2=12μ F,V=150volt . Total capacity, (1/C)=(1/C1)+(1/C2)=(1/6)+(1/12) (1/C)=(3/12)⇒ C=4μ F Potential of 12μ F capacitor V=(q/C) V= (4 × 150/12) V=50V

Q. Two capacitors of capacity $6 μ F$ and $12 μ F$ in series are connected by potential of $150 V$ . The potential of the capacitor of capacity $12 μ F$ will be

$25 V$

$50 V$

$100 V$

$150 V$

Given, $C_{1} = 6 μ F, C_{2} = 12 μ F, V = 150 v o lt$ .
Total capacity, $\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} = \frac{1}{6} + \frac{1}{12}$
$\frac{1}{C} = \frac{3}{12} \Rightarrow C = 4 μ F$
Potential of $12 μ F$ capacitor
$V = \frac{q}{C}$
$V = \frac{4 \times 150}{12}$
$V = 50 V$

Q. Two capacitors of capacity 6μF and 12μF in series are connected by potential of 150V . The potential of the capacitor of capacity 12μF will be

25V

50V

100V

150V