First slide

Capacitance

Question

Two capacitors A and B are connected in series with a battery as shown in fig. When the switch S is closed and the capacitors get charged fully, then

Easy

A

the p.d. across the plates of A is 4 V and across the plates of B is 4 V
B

the p.d. across the plates of A is 6 V and across the plates of I is 4 V
C

the ratio of electrical energies stored in A and B is 2:3
D

the ratio of charges on A and B is 3 : 2

Solution

The two capacitors are connected in series. The equivalent capacitance C is given by
$\begin{array}{l} \frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \\ C = 6 / 5 μF \end{array}$
Charge on each capacitor =6 / 5 x 10 = 12 coulomb.
Potential diff. across $A = \frac{12}{2} = 6 volt$
Potential diff. across $B = \frac{12}{3} = 4 volt$

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