A7.00 µF capacitor is connected to a 12.5 mH inductor as shown in the circuit above. What happens to the current when the switch is closed? O No currrent flows in the circuit. O The current will begin to oscillate in the circuit. O The current flows and is disspated in the inductor.

I'm working on some homework and am stumped with this problem.  Help would be great, thanks!

(a) What is the current through the inductor immediately after the switch is closed? (b) Write symbolic expressions for the current through resistors R and 2R immediately after the switch is closed, in terms of R, &, and L, as needed, plus constants, then calculate each numeric value. 11₁ = 12 = 11 A A

The figure below shows a circuit with a battery of EMF = 3.75 V, a resistor with R = 2.16 2, and an inductor with L = 2.45 H. If the battery is connected at t = 0, at what time will the rate of energy stored in the inductor be equal to x = 1.89 times the power dissipated in the resistor? i S 99 4+ R L2 2 Hint You can use the conservation of energy to determine the fraction of the power dissipated in the resistor: total energy lost by a battery per unit of time (power) is &I is spent on the rate of energy stored by the inductor and the power dissipated in the resistor I² R. As EMF, resistance, and inductance are given to you, you know the time dependence of the current to calculate the fraction x at any given time.