Wheatstone Bridge and Metre Bridge

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. We want to measure an unknown resistance, and we use a galvanometer, a meter that detects current. But we never read a number from it; we just wait until it shows zero. So how can a meter showing nothing tell us a resistance? And stranger still, the answer does not depend on the battery at all. That is the Wheatstone bridge and the metre bridge, and in JEE, questions on them come up often. So, let's begin.

0:32Wheatstone bridge: zero current in the middle

So, here is the Wheatstone bridge on your screen. Four resistors, P, Q, R and S, form a diamond with corners A, B, C and D, as you see. The cell, with e m f capital E, is joined across A and C, and the middle arm, B D, carries G, the galvanometer. Okay? Now, when does no current flow through G? Read the note beside the figure: the bridge is balanced when P over Q equals R over S, the line now on screen. Then B and D are at the same potential; V B equals V D, where V with a small B or D below is the potential at that corner. So the galvanometer current, I G, that is I with a small G below, is zero. And since no current flows in B D, the note says that arm may even be removed, or shorted, and nothing changes. Clear? Now, a quick test for you. The bridge is balanced, and we double the e m f of the cell. Does the balance break? No, the balance stays exactly as it was! Look at the condition: P over Q equals R over S, with no E and no G in it. So the balance does not depend on the e m f, or on the galvanometer. Got it? And one more surprise. Swap the battery and the galvanometer, and the bridge still stays balanced. Remember this one. With me?

2:01Metre bridge: a uniform wire makes two arms

Now, the metre bridge, which is a Wheatstone bridge too. In its left gap we put the unknown resistance X, and in the right gap a known resistance R, okay? The other two arms come from one uniform wire, one hundred centimetres long, that is one metre, hence the name. Its two parts are the other two arms. Right? The point on the wire where the galvanometer reads zero is the null point. Say it is at a distance l from the left end. Then the two parts have lengths l and one hundred minus l, in centimetres. With me? Now, the balance condition needs the resistances of these two parts, but we only know their lengths. So why are lengths enough here? Yes, exactly, because the wire is uniform! Same material and same thickness all along, so each part's resistance is proportional to its length. So the balance condition becomes X over R equals l over one hundred minus l. The unknown in the left gap pairs with l, the length on the left. Got it?

3:05What moves the null point?

Now, a common trap. We replace the wire with a thicker one, still uniform. Its resistance changes, right? So does the null point move? Many students say yes, but no, it stays exactly where it was! Only the ratio of the two parts matters, and for a uniform wire that is l over one hundred minus l, whatever the thickness. And a different cell does not move it either. Clear? But if the wire is not uniform, resistance is no longer proportional to length, and the null point moves. Okay?

3:39End corrections and the best null point

Now, a real metre bridge is not perfect at the two ends of the wire. We count this as extra effective length: alpha at the left end and beta at the right end, the two Greek letters on screen. These are the end corrections. Got it? So the balance condition becomes X over R equals l plus alpha, over one hundred minus l plus beta. Each end gets its own correction. Right? In practice, we interchange X and R, take the reading again, and average. This reduces the errors. With me? Now the error. Delta X, with the little triangle, is the error in X, and delta l, the error in reading the null point. The relative error, delta X over X, equals one hundred times delta l, over l times the bracket one hundred minus l. So where should the null point be, to keep this error smallest? Yes, exactly, near the middle of the wire! The bottom of the fraction, l times one hundred minus l, is largest when l is fifty, so the error is least there. Got it?

4:47Example 1: null point at 40 cm with 3 ohm, find X

Alright, example one. In a metre bridge, the null point is at forty centimetres, with X in the left gap and three ohm in the right; the Greek letter omega is the symbol for ohm. Find X, okay? Pause the video and try it first. Done? Okay, let's check. Many students slip here. X is in the left gap, so does X go with forty, or with sixty? It goes with forty, of course! The null point is measured from the left end, so the left gap pairs with l, which is forty. The right gap pairs with one hundred minus forty, which is sixty. So, X over three equals forty over sixty. Right? Now multiply both sides by three. X equals three times forty over sixty, which gives two ohm. So keep the left gap paired with l, and you will not fall into that trap. Clear? And here is a quick check that toppers use. The null point is left of the middle, so the left resistance must be smaller; the sign on screen means less than. Two is less than three, so the answer makes sense. Got it?

5:59Example 2: X shunted by 12 ohm, null point moves to 50 cm

Now a harder one. With X in the left gap, the null point is at sixty centimetres. When X is shunted by twelve ohm, the null point moves to fifty centimetres. Find X, and R, the resistance in the right gap. Pause the video and try it first. Done? Okay, let's check. Before the shunt, X over R equals sixty over forty, which is one point five. The double arrow on screen means this gives, so X equals one point five times R, okay? Now, the key word here is shunted. What does it mean for X? Yes, exactly, a shunt is connected in parallel! So the left gap now holds X in parallel with twelve ohm; the two vertical lines mean in parallel with. The null point is at fifty, the middle, so the ratio is one, and X in parallel with twelve equals R itself. Right? Now write both sides using X only. X in parallel with twelve is product over sum: twelve X over X plus twelve. And R is X over one point five, from the first line. So, twelve X, over X plus twelve, equals X over one point five. With me? X is not zero, so cancel X on both sides, and cross multiply. Twelve times one point five equals X plus twelve. Twelve times one point five is eighteen, so X equals eighteen minus twelve, which gives six ohm. Clear? And R is six over one point five, which gives four ohm. Got it? Let us check. Six in parallel with twelve is seventy two over eighteen, which gives four ohm, exactly the value of R, as it should be. And see, the shunt lowered the left resistance, so the null point moved left. Makes sense, right?

7:55Quick revision: P/Q = R/S and X/R = l/(100 - l)

So, quick revision. Balanced Wheatstone bridge: P over Q equals R over S, and no current flows through the galvanometer. Metre bridge: X over R equals l over one hundred minus l, and the unknown in the left gap pairs with l. And for the least error, keep the null point near the middle. Now go and try the practice problem on this page. All the best!