CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
C2N21N229
loader
C
C2N19N2
loader
D
C2N19N1
loader

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).

Physics
NCERT
Standard XII

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image