Question

A capacitor of capacitance $2\mu$ F is charged to a potential difference of 12 V. It is then connected across an inductor of inductance 0.6 mH. The current in the circuit when the potential difference across the capacitor is 6 V is :

• A. 3.6 A
• B. 2.4 A
• C. 1.2 A
• D. 0.6 A

Answer 1

The correct option is D 0.6 A

According to question,

Initial energy of capacitor $E_i=\frac{1}{2}C{V}_1^2$

Final energy of capacitor is $E_f=\frac{1}{2}C{V}_2^2$

Final energy of inductor $E_L=\frac{1}{2L{I}^2}$

The total energy in the given circuit remains conserved I.e.

$\frac{1}{2}C{V}_1^2=\frac{1}{2}C{V}_2^2+\frac{1}{2}L{I}^2$

$C(V_1^2-V_2^2)=L{I}^2$

$I=\sqrt{\frac{C(V_1^2-V_2^2)}{L}}$

Given that,

$C=2\mu F=2\times {10}^{-6}F$

$L=0.6\times {10}^{-3}H$

$V_1=12V$

$V_2=6V$

Put all values in the expression of I

$I=\sqrt{\frac{2\times {10}^{-6}(144-36)}{0.6\times {10}^{-3}}}$

$I=\sqrt{\frac{2\times {10}^{-3}\times 108}{0.6}}$

$I=0.6A$