The difference in the effective capacity of two similar capacitors when joined in series and then in parallel is 6 μF. The capacity of each capacitor is

Question

The difference in the effective capacity of two similar capacitors when joined in series and then in parallel is 6 μF. The capacity of each capacitor is (A) 2 μF (B) 4 μF (C) 8 μF (D) 16 μF

Tardigrade Admin · Accepted Answer

When two similar capacitors are joined in series, then effective capacity is

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

∴ C_{s} = \frac{C}{2}

When capacitors are joined in parallel, then

C_{p} = C + C = 2 C

Since,

C_{p} - C_{s} = 6 μ F

∴ 2 C - \frac{C}{2} = 6 \Rightarrow 4 C - C = 12

or

C = 4 μ F

Q. The difference in the effective capacity of two similar capacitors when joined in series and then in parallel is $6 μ F$ . The capacity of each capacitor is

$2 μ F$

$4 μ F$

$8 μ F$

$16 μ F$

Q. The difference in the effective capacity of two similar capacitors when joined in series and then in parallel is 6μF. The capacity of each capacitor is

2μF

4μF

8μF

16μF

When two similar capacitors are joined in series, then effective capacity is Cs​1​=C1​+C1​=C2​ ∴Cs​=2C​ When capacitors are joined in parallel, then Cp​=C+C=2C Since, Cp​−Cs​=6μF ∴2C−2C​=6⇒4C−C=12 or C=4μF

Q. The difference in the effective capacity of two similar capacitors when joined in series and then in parallel is $6 μ F$ . The capacity of each capacitor is

$2 μ F$

$4 μ F$

$8 μ F$

$16 μ F$