LR Circuits

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. With only a resistor, the current reaches its full value the moment you close the switch. Now put a big coil in series and close the switch again. Does the current still jump to its full value at once? It does not. It creeps up, like a heavy truck picking up speed. One number, the time constant, tells us how slowly. LR circuits come up often in JEE, so let's begin.

0:30Growth of current: the coil slows the rise

Alright, let's build the circuit. Here is a coil of self-inductance L, a resistor R, and a battery whose emf is a curly E, connected at time t equal to zero. The graphs on the right show the current, small i, against time, small t, okay? Now tell me. Just after we close the switch, what is the current in the circuit? It starts from zero, yes, exactly! By Lenz's law, the coil's back emf opposes the rise, so the current cannot jump; it grows. The current is curly E over R, times one minus e to the power minus t over tau, where e is the base of natural logarithms. It settles at i nought, the i with a small zero below, equal to curly E over R, the top dashed line in the left graph. Right? And tau, the Greek letter, is the time constant, L over R, okay? Henry over ohm gives seconds, so tau really is a time. So a bigger L means a longer tau and a slower rise. Clear? Now the voltage across the coil, V with a small L below. It equals curly E times e to the power minus t over tau. At time zero it is the whole emf, because no current flows yet, so nothing drops across the resistor. Then it dies away. With me? Here is a number worth remembering. At t equal to tau, the current is i nought times one minus one over e, and the wavy equals sign means approximately, so about zero point six three times i nought. That is sixty three percent of the final value, as the dashed lines on the graph show. Got it?

2:08Just after a switch, and long after

Now two rules that crack most switch questions. Rule number one: the current through an inductor cannot change suddenly. So just after a switch is thrown, the inductor keeps the current it had before. Okay? And if that earlier current was zero, it stays zero just after the switch. The double arrow means which gives: the inductor acts as an open circuit, like a break in the wire. Right? Rule number two is about the steady state, a long time later. The current has stopped changing, so d i over d t, the rate of change of current, is zero. Clear? So, with d i over d t equal to zero, what does an ideal inductor act like? Like a plain wire, yes, exactly! The back emf, L times d i over d t, is zero, and an ideal inductor has no resistance. So remember the pair: an open circuit at the start, if it carried no current before, and a plain wire in the steady state. Got it?

3:09Decay, and the time to reach a fraction

Now the reverse. We remove the battery and close the circuit on itself. The current does not stop at once, because by Lenz's law the back emf now acts along the current, trying to keep it going. Look at the falling curve on the right. Okay? The current dies away: i equals i nought times e to the power minus t over tau. And at t equal to tau, only about zero point three seven times i nought is left. Right? Back to growth now. How long does the current take to reach a fraction f of its final value? Set one minus e to the power minus t over tau equal to f, and solve. We get t equals tau times l n, the natural log, of one over one minus f, clear? So tell me, how long does the current take to reach half of its final value? Put f equal to one half, and one over one minus f becomes two. So t equals tau times the natural log of two, which is zero point six nine three times tau. With me? And for ninety percent, the percent sign on screen, one over one minus f is ten. So t equals tau times the natural log of ten, which is two point three zero times tau, more than two time constants. Got it?

4:25The trap: which R goes into the time constant?

Now a classic trap. Suppose a switch changes the circuit, so the inductor current now flows through a different path. Is the time constant still the same as before? No, careful, it can change! Tau equals L over R uses the resistance of the path through which the inductor current actually flows at that stage. Okay? So whenever a switch changes the circuit, find the new resistance for the new path, and from it, the new tau. Right? And across the switch, the inductor current stays continuous: it starts the new stage with exactly the value it had at that moment. Clear?

5:03Example 1: time constant and current of a 3 H, 6 ohm coil

Now our first example. Here is a coil of three henry and six ohms, connected to a twelve volt battery. On screen, capital H is henry, capital omega is ohm, V is volt, and s is second. Find the time constant, the final current, and the current after zero point five seconds. Pause the video and try it first. Done? Okay, let's check. Time constant first: tau equals L over R, so three over six, which gives zero point five seconds. With me? The final current is i nought, curly E over R, so twelve over six gives two amperes, with capital A for ampere. Right? Now the current after zero point five seconds. Is there something special about that time? Yes, exactly, zero point five seconds is exactly one time constant! So i equals two times one minus one over e, about zero point six three times two, which gives one point two six amperes. Okay?

6:06Example 2: battery current just after and long after closing

Now a harder switch question. Here a ten volt battery is connected through R one, two ohms, to an ideal inductor in parallel with R two, three ohms. R one and R two are R with a small one and a small two below. Find the battery current just after the switch is closed, and a long time later. Pause the video and try it first. Done? Okay, let's check. Many students say the inductor is a wire, so it shorts R two, and the current is ten over two, five amperes, straight away. Is that right? No, no, careful, that is the trap! Just after closing, the inductor current cannot jump from zero, so the inductor is an open circuit. Got it? So the current flows through R one and R two in series. Ten over two plus three, that is ten over five, gives two amperes. Clear? Now a long time later. The current is steady, so the ideal inductor is a plain wire, and it short-circuits R two, okay? Only R one is left in the path. Ten over two gives five amperes. So five amperes was correct, but only in the steady state. With me?

7:18Quick revision: growth, decay, tau = L/R, which R to use

So, quick revision. First, growth is i nought times one minus e to the power minus t over tau, decay is i nought times e to the power minus t over tau, and tau is L over R, okay? Second, just after a switch, the inductor keeps its current, and in the steady state it is a plain wire. Third, tau uses the resistance of the path the inductor current flows through. Now go and try the practice problems on this page. All the best!