Question

# A series combination of $$N_1$$ capacitors (each of capacity $$C_1$$) is charged to potential difference $$'3V'$$. Another parallel combination of $$N_2$$ capacitors (each of capacity $$C_2$$) is charged to potential difference 'V'. The total energy stored in both the combinations is same. The value of $$C_1$$ in terms of $$C_2$$ is

A
C2N1N29
B
C2N21N229
C
C2N19N2
D
C2N19N1

Solution

## The correct option is A $$\dfrac{C_2N_1N_2}{9}$$For series combination:-equivalence capacitance=$$\dfrac{C_1}{N_1}$$Hence, charge on each capacitor, $$q=(3V)\dfrac{C_1}{N_1}$$Hence, total energy, $$U=N_1\times \dfrac{q^2}{2C_1}=\dfrac{9C_1V^2}{2N_1}$$For parallel combination:-Potential across each capacitor is $$V$$.Hence, total energy, $$U=N_2\times \dfrac{1}{2}C_2V^2=\dfrac{N_2C_2V^2}{2}$$$$\dfrac{9C_1 V^2}{2N_1}=\dfrac{N_2 C_2 V^2}{2}$$$$\implies C_1=\dfrac{N_1N_2C_2}{9}$$Hence, answer is option-(A).PhysicsNCERTStandard XII

Suggest Corrections
Â
0

Similar questions
View More

Same exercise questions
View More

People also searched for
View More