Four identical capacitors are connected as shown in diagram. When a battery of 6 V is connected between A and B, the charge stored is found to be 1.5 μ C. The value of C1 is

Question

Four identical capacitors are connected as shown in diagram. When a battery of 6 V is connected between A and B, the charge stored is found to be 1.5 μ C. The value of C1 is (A) 2.5 μ F (B) 15 μ F (C) 1.5 μ F (D) 0.1 μ F

Tardigrade Admin · Accepted Answer

The capacitance across A and B =(C1/2)+C1+C1=(5/2) C1 As Q=C V, 1.5 μ C =(5/2) C1 × 6 ⇒ C1=(1.5/15) × 10-6 =0.1 × 10-6 F =0.1 μ F

Q. Four identical capacitors are connected as shown in diagram. When a battery of $6 V$ is connected between $A$ and $B$ , the charge stored is found to be $1.5 μ C$ . The value of $C_{1}$ is

$2.5 μ F$

$15 μ F$

$1.5 μ F$

$0.1 μ F$

The capacitance across $A$ and $B$
$= \frac{C _{1}}{2} + C_{1} + C_{1} = \frac{5}{2} C_{1}$
As $Q = C V$ ,
$1.5 μ C = \frac{5}{2} C_{1} \times 6$
$\Rightarrow C_{1} = \frac{1.5}{15} \times 1 0^{- 6}$
$= 0.1 \times 1 0^{- 6} F = 0.1 μ F$

Q. Four identical capacitors are connected as shown in diagram. When a battery of 6V is connected between A and B, the charge stored is found to be 1.5μC. The value of C1​ is

2.5μF

15μF

1.5μF

0.1μF