A 3 H inductor with a resistance of 7 ohms is connected to a 12 V battery with negligible internal resistance. What is the current through the inductor 0.5 seconds after the battery is connected?
At t0, switch closed and a potential came across the inductor. So it is like a Alternating current for a Moment. That alternating current makes inductor to behave like in AC circuit. That affects reduce continuously after t+ time because of dc voltage. There will a opposing electromagnetic field in the inductor due to sudden voltage. that will stop current to flow through the inductor
Consider the inductor with its internal resistance to be like a perfect inductor (having zero resistance) with a resistor in series (many people seem to prefer this mental picture). This means the total voltage across the inductor and resistor equals the sum of the voltages across each. Use the inductor and resistor equations at Wikipedia (see url below) to get the total voltage across the inductor and resistor. V(total)V(inductor)+V(resistor) V(total)Ldi/dt + iR Solve the differential equation for i as a function of t, remembering the initial condition that i0 at time0.
