Mixed Problems: Set 2 (Cells, Kirchhoff's Laws and Instruments)

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Join a two volt cell to a resistor, and put an ideal voltmeter across the cell. What does it read? Two volts, right? Many students say yes, and lose an easy mark. Keep your answer in mind. Today is our second set of mixed problems, on cells, Kirchhoff's laws and instruments, and questions like these come up often in JEE Main. So, let's begin.

0:29Idea 1: a real cell loses voltage inside

Let us begin with idea number one, our puzzle in numbers. Here, a cell of two volts, with an internal resistance of zero point five ohm, drives an external three point five ohm. Capital V means volt, and the Greek letter omega means ohm. Find the current and the terminal voltage. So, will a voltmeter across the cell read the full two volts? Let us work it out and see. Inside every real cell sits an internal resistance, small r, in series with capital R, the outside resistance. So the current I equals E, the emf, over capital R plus small r. Two over three point five plus zero point five is two over four, which gives zero point five ampere; capital A means ampere. Okay? Now, while the cell gives out this current, part of its emf is used up inside it, across small r. So the terminal voltage, V, equals E minus I times small r, right? Put in the numbers: two minus zero point five times zero point five, where the cross means times. That gives one point seven five volts, not two volts. So the voltmeter reads less, and the missing quarter of a volt is lost inside the cell. That solves our puzzle. Clear? And one more case. When the cell is being charged, current is pushed into its positive terminal, so the sign flips, and V equals E plus I times small r, because the charger must also push the current through small r. Minus while it discharges, plus while it is charged; students often mix these up. Got it?

2:05Idea 2: a galvanometer becomes a meter

Now idea number two, the meters. Here, a galvanometer of twenty ohm reads full scale at ten milliampere, written m A on screen. Find the shunt for a two ampere ammeter, and then turn the same galvanometer into a thirty volt voltmeter. In the figure, the galvanometer is the capital G in a circle, and capital G is also its resistance. Okay? The current above the circle, I g, that is I with a small g below, is the full-scale current. Ten milliampere is zero point zero one ampere; always change to amperes first. With me? On the left of the figure, the arrow shows the big current I coming in; most of it bypasses the coil through a small shunt, capital S, in parallel. So S equals I g times G, over I minus I g, as written under the figure. That is zero point zero one times twenty, over one point nine nine, where one point nine nine is two minus zero point zero one. That comes to about zero point one ohm; the wavy equals sign means approximately. Clear? Now the voltmeter, on the right of the figure: a large resistance, capital R, in series with the galvanometer. At thirty volts, a current of zero point zero one ampere must flow. So the series resistance is thirty over zero point zero one, which is three thousand ohm. Is that it? No, no, careful, three thousand ohm is the total resistance we need, and the galvanometer already gives twenty ohm of it. So R equals V over I g, minus G, as written under the figure, which here is thirty over zero point zero one, minus twenty. Okay? Three thousand minus twenty gives two thousand nine hundred and eighty ohm, joined in series. Remember the pair: a small shunt in parallel for an ammeter, and a large resistance in series for a voltmeter.

3:57Example 1: potentiometer, compare emfs at 45 cm and 60 cm

Got it? Alright, example one, the potentiometer. One cell, of one point five volts, balances at forty five centimetres, and another cell at sixty centimetres. Find the second emf. Pause the video and try it first. Done? Okay, let's check with the figure. At the top, the driver cell, epsilon naught, the Greek letter epsilon with a small zero below, sends a steady current through a rheostat and the uniform wire from A to B, so the potential falls evenly along the wire. Below, the cell E and the galvanometer run from A to the jockey J, a sliding contact. With me? The drop per unit length is k, the potential gradient: VAB, the V with a small A B below, is the voltage across the wire, and capital L is its length. At the null point the galvanometer reads zero, so no current is drawn from the cell, and E equals k times small l, the balancing length that the double arrow just above the wire marks. Okay? Now, the second cell needs sixty centimetres instead of forty five. With E equal to k times small l, is its emf bigger or smaller? It is bigger, that is correct. Both cells use the same wire and the same driver, so k is the same, and E is proportional to small l. So E two over E one equals l two over l one, where the small one and two below mark the two cells. Clear? So E two equals one point five times sixty over forty five. One point five times sixty is ninety, and ninety over forty five gives two volts. Flip the ratio by mistake and you get less than one point five, which cannot be right. Got it?

5:43Example 2: junction potential from 12 V, 0 V and 6 V points

Now a harder one, on Kirchhoff's laws. From a junction O, two ohm goes to a point at twelve volts, four ohm to a point at zero volts, and four ohm to a point at six volts. Find V O, the capital V with a small O below, which is the potential of the junction. Pause the video and try it first. Done? Okay, let's check. Many students take the average of twelve, zero and six, and say six volts. Is that correct? No, no, wait, the resistances are not equal, so a plain average cannot work. We use Kirchhoff's junction rule: the current flowing into a junction equals the current flowing out, because charge is conserved. Okay? Now our sign convention, stated once. Every current is taken as flowing into O, towards the junction. From a point at potential V one, the V with a small one below, the current through a resistance R is V one minus V O, over R, as on screen. If any comes out negative, it simply flows out. So these currents add up to zero. Right? So, twelve minus V O, over two, plus zero minus V O, over four, plus six minus V O, over four, equals zero. With me? Multiply every term by four to clear the fractions. Twenty four minus two V O, minus V O, plus six minus V O, equals zero. The double arrow means gives, so we get thirty equals four V O, clear? So V O is thirty over four, which gives seven point five volts, not six. It is pulled up towards twelve volts, because the two ohm link to that point has the smallest resistance. Very good, well done!

7:26Quick revision: V = E - Ir, shunt, E = kl, junction rule

So, quick revision. For a real cell, I is E over capital R plus small r, and the terminal voltage is E minus I times r while it discharges, and E plus I times r while it is charged. An ammeter needs a small shunt in parallel, and a voltmeter needs V over I g, minus G, in series. The potentiometer gives E equals k times l, drawing no current, so emfs are in the ratio of their balancing lengths. And at a junction, take every current into the node; they add up to zero. Now go and try the practice problem on this page. All the best!