Parallel Branches on AC

Transcript

0:00Introduction

Hello everyone! Here is a strange little puzzle. An ideal inductor and an ideal capacitor are connected in parallel across an A C source, at one special frequency. Each of them carries a large current. But the ammeter at the source reads zero! So where does all that current come from? To answer this, we need to see how currents add in parallel branches. This idea does come up in JEE questions. So, let's begin.

0:30Parallel branches: same voltage, different currents

So, first the basic rule. Branches in parallel are connected across the same two points, so every branch has the same voltage, capital V, okay? In series the current was common; in parallel, the voltage is common. Right? Now each branch takes its own current. I R, capital I with a small R below, is the resistor's current, V over R, okay? I L is V over X L, the inductor's reactance, and I C is V over X C, the capacitor's reactance. Clear? But these currents are not in step. Draw the voltage along the horizontal. The resistor's current lies along it, in phase with the voltage. Okay? The inductor's current is ninety degrees behind the voltage; the small circle means degrees. The capacitor's current is ninety degrees ahead. Current lags in an inductor and leads in a capacitor, as you know. With me? Now, let me ask you this. Is the total current from the source just I R plus I L plus I C, like in a D C circuit? No, no, careful, that is the trap! These currents point in different directions, so they add as phasors, like vectors, not as plain numbers. Got it?

1:42Adding the branch currents

So, let us add them. I L is ninety degrees behind and I C is ninety degrees ahead, so they point in opposite directions; we say they are in antiphase. Together they give I C minus I L, okay? And I R is at ninety degrees to both of them. So how do we add two phasors at right angles? With Pythagoras, yes, exactly! The total current, capital I, is the square root of, I R squared, plus I C minus I L, the whole squared. This one line is the heart of today's lesson. Clear? Notice how it matches the series circuit. In series, the current is common and the voltages add this way; in parallel, the voltage is common and the currents add this way.

2:26An ideal inductor and capacitor in parallel

Right? Now, back to our puzzle. Take an ideal inductor and capacitor in parallel, with no resistor, so I R is zero. Then I equals mod of I C minus I L; the two straight bars mean the modulus, the size without the sign. Okay? Omega is the angular frequency. At omega equals one over the square root of L times C, the reactances X L and X C become equal. Same voltage, same reactance, so I L equals I C as well. Right? So, what current does the source supply now? The answer is zero, absolutely! The two equal currents are in antiphase, so they cancel completely. No current for a finite voltage means the impedance, V over I, is infinite. Got it? But the inductor and capacitor still carry their large currents. That current simply circulates between L and C, while the source supplies nothing. So that is how our ammeter read zero! With me?

3:28Example 1: 30 ohm R and 40 ohm inductor in parallel on 120 V

Now, example one. We have a resistor of thirty ohms and an inductor of reactance forty ohms in parallel across one hundred and twenty volts A C, okay? The capital omega is ohms, and capital V is volts. Find the total current, the impedance and the power factor. Pause the video and try it first. Done? Okay, let's check. Both branches get the full one hundred and twenty volts. So I R is one hundred and twenty over thirty, which gives four amperes, written capital A, in phase with the voltage. Okay? And I L is one hundred and twenty over forty, which gives three amperes, lagging by ninety degrees. Right? There is no capacitor, so I is the square root of four squared plus three squared, which gives five amperes. And it is not seven, the plain sum, which is the trap we just saw. Clear? Now the impedance. Many students write the square root of thirty squared plus forty squared, which gives fifty ohms. Is that right here? No, no, wait, that formula is for a series circuit. Impedance always means V over I, so Z is one hundred and twenty over five, which gives twenty four ohms. Got it? Finally, the power factor, cos phi, where phi is the phase angle between the current and the voltage. Only I R is in phase with the voltage, so cos phi is I R over I, four over five, which gives zero point eight. With me?

4:57Example 2: RC branch and RL branch in parallel on 20 V rms

Now a harder one, JEE level. One branch has R one, R with a small one below, equal to one hundred ohms, in series with a capacitor of reactance one hundred ohms. In parallel, a second branch has R two, fifty ohms, in series with an inductor of reactance fifty ohms. The source is twenty volts r m s, the root mean square value. Find the total current. Pause the video and try it first. Done? Okay, let's check. Each branch is a small series circuit. For branch one, Z one is the square root of one hundred squared plus one hundred squared, which gives one hundred root two ohms. And tan phi one, the reactance over the resistance, is one. Okay? So I one is twenty over one hundred root two, which is zero point two over root two amperes, an r m s current. Tan phi one is one, so phi one is forty five degrees, and in a capacitor branch the current leads. Right? Branch two is the same idea: Z two is the square root of fifty squared plus fifty squared, which gives fifty root two ohms, and tan phi two is one. Clear? So I two is twenty over fifty root two, which is zero point four over root two amperes. This branch has the inductor, so its current lags by forty five degrees. With me? Now, the total current. One current leads by forty five degrees, the other lags by forty five degrees. Can we just add them? And if not, what is the angle between them? No, we cannot simply add them, because they are forty five plus forty five, that is ninety degrees apart! So, Pythagoras again: I is the square root of I one squared plus I two squared. I one squared is zero point zero two, and I two squared is zero point zero eight. Okay? So I is the square root of zero point one, which is about zero point three two amperes; the wavy equals sign means approximately. One branch at a time, and it is easy, right?

7:00Quick revision: phasor sum, Z = V/I, power factor, resonance

So, quick revision. In parallel, every branch has the same voltage, and the branch currents add as phasors. The total current is the square root of I R squared plus I C minus I L, the whole squared; the impedance is V over I, and the power factor is I R over the total current. And an ideal inductor and capacitor in parallel, at omega equals one over the square root of L times C, draw no source current, so the impedance is infinite. Now go and try the practice problems on this page. All the best!