Question

An inductor of inductance 2.0 mH is connect across a charge capacitor of capacittance 5.0 uF , and the resulting LC circuit is set oscillating at natural frequency. Let Q denote the instantaneous charge on the capacitor, and I the current in the circuit. It is found the maximum value of Q is 200 uC .When Q=100uC the value of $\frac{dI}{dt}$ is found to be ${10}^x$ A/s. Find the value of x is then

• A. x=-4
• B. x=-5
• C. x=4
• D. x=5

Answer 1

The correct option is A x=-4

In an $Le$ system $he$ charge oscillation between $I$ and $Q$

Applying $KVL$ gives,

$-L\frac{dI}{dt}+\frac{9}{c}=0$ where $L=2mH$

$c=5\mu F$

also, maximum charge $Q_0=200\mu c$

Thus, total energy of system, $\mu =\frac{Q_0}{2c}=\frac{200Z^2}{2\times 5}=4000J$

when $9=100uc$, we get from $\left(1\right)$,

$\frac{dI}{dt}dfrac+9Lc=\frac{+100}{2\times {10}^{-3}\times 5}=+{10}^{-4}A/s$

as $\frac{dI}{dt}={10}^x$ in gives $x=-4$