A 3.0-$\mu$F capacitor was fully charged by being connected to a 9.0-V battery.The fully charged capacitor is then connected to a 0.50-H inductor.The maximum current in the inductor is
A)22 mA.
B)0.054 mA.
C)10.2 mA.
D)0.028 mA.
E)35 mA.

Question

Quizplus · Accepted Answer

The maximum current in the inductor can be found using the conservation of energy principle. The energy stored in the capacitor when fully charged is equal to the energy in the inductor at maximum current. The energy in the capacitor is given by $ \frac{1}{2}CV^2 $, and the energy in the inductor is given by $ \frac{1}{2}LI^2 $, where $ C $ is the capacitance, $ V $ is the voltage, $ L $ is the inductance, and $ I $ is the current. Setting these equal to each other and solving for $ I $ gives:$$ \frac{1}{2}CV^2 = \frac{1}{2}LI^2 $$Given $ C = 3.0 \mu F = 3.0 \times 10^{-6} F $ and $ L = 0.50 H $, and $ V = 9.0 V $, we can solve for $ I $:$$ 3.0 \times 10^{-6} F \times (9.0 V)^2 = 0.50 H \times I^2 $$$$ I = \sqrt{\frac{3.0 \times 10^{-6} F \times (9.0 V)^2}{0.50 H}} $$$$ I = \sqrt{\frac{3.0 \times 10^{-6} \times 81}{0.50}} $$$$ I = \sqrt{\frac{243 \times 10^{-6}}{0.50}} $$$$ I = \sqrt{486 \times 10^{-6}} $$$$ I = 22 \times 10^{-3} A $$$$ I = 22 mA $$Therefore, the maximum current in the inductor is 22 mA.

A 30- μ\muμ F Capacitor Was Fully Charged by Being Connected to a Connected

A 30- $\mu$ F Capacitor Was Fully Charged by Being Connected to a Connected