Question

A series combination of 5 capacitor, each of capacitance 1 microfarad, is charged by a source of potential difference 2V. When another parallel combination of 10 capacitor each of capacitance C2, is charged by a source of potential difference V, it has the same total energy stored in it as the first combination has. the value of C2 is​

Answer 1

Answer

• Option C
• 2/25 μF

Given

• A series combination of 5 capacitor, each of capacitance 1 microfarad, is charged by a source of potential difference 2V. When another parallel combination of 10 capacitor each of capacitance C2, is charged by a source of potential difference V, it has the same total energy stored in it as the first combination has

To Find

• Value of C₂

Formula Used

$\bullet \;\;\; \rm U=\dfrac{1}{2}C_{eq}V^2$

When  capacitors of same capacitance are connected in series ,

$\bullet \;\;\; \rm C_{series}=\dfrac{C}{n}$

When capacitors of same capacitance are connected in parallel ,

$\bullet \;\;\; \rm C_{parallel}=nC$

Solution

Case - 1 :

5 capacitors each of capacitance 1 microfarad are connected in series .

$\rm C_{series}=5(1 \mu)\\\\\implies \rm C_{series}=\dfrac{\mu}{5}$

Calculate energy .

Potential difference , V = 2 V

$\implies \rm U_{1}=\dfrac{1}{2}C_{series}(V)^2\\\\\implies \rm U_1=\dfrac{1}{2} \times \dfrac{\mu}{5} \times 2^2\\\\\implies \rm U_1=\dfrac{2\mu}{5}$

Case - 2 :

Now another parallel combination of 10 capacitor each of capacitance C₂

$\implies \rm C_{parallel}=10C_2$

Potential difference , V = 1 V

Calculate Energy

$\implies \rm U_{2}=\dfrac{1}{2} C_{parallel}V^2 \\\\\implies \rm U_2=\dfrac{1}{2} \times 10 C_2 \times 1^2\\\\\implies \rm U_2=5C_2$

A/c , " it has the same total energy stored in it as the first combination has "

$\implies \rm U_1=U_2\\\\\implies \rm \dfrac{2\mu}{5} =5C_2\\\\\implies \rm C_2=\dfrac{2\mu}{25}\\\\\implies \rm C_2=\dfrac{2}{25} \mu F$

Answer 2

Answer:

2/25

Explanation:

answer is 0.08 = 2/25