Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You charge a capacitor from a cell through a resistor, and the resistor gets warm. Now swap in a much bigger resistor. More resistance, so more heat, right? Hold on to that answer. Today we work through the six concept checks of this unit's problem bank. Statement questions like these come up often in JEE, so let's begin.

0:27Check 1: does every conductor obey Ohm's law?

So, check one. Ohm's law says V equals I times R, where V is the potential difference, I the current, and R the resistance. It holds for a metallic conductor at constant temperature. Okay? Now here is the check. Does every conductor obey Ohm's law? First, what does obeying it mean? For an ohmic conductor, the V I graph, V against I, is a straight line through the origin. With me? So the answer is no. Diodes, gas discharges, and filament lamps over a wide range of current do not give such a straight line, so they are non-ohmic. Clear?

1:08Check 2: why manganin for standard resistors?

Now check two. For a metal, R sub T equals R nought times, open bracket, one plus alpha times delta T, close bracket. R nought, with a small zero, is the resistance before heating, R sub T, with a small T, the resistance after, and delta T, with delta the Greek triangle, the rise in temperature. Alpha, the Greek letter, is the temperature coefficient of resistance; for copper, about four times ten to the power minus three per kelvin; the wavy equals sign means approximately, and K to the power minus one means per kelvin. Okay? But a standard resistor must keep its value. So why is it made of manganin, an alloy? First, manganin has a very small alpha, so its resistance hardly changes with temperature. With me? And it has a high resistivity, rho, the Greek letter like a curly p. From R equals rho times l, over capital A, with l the length and capital A the area of cross-section, a short wire then gives a large resistance. Got it?

2:14Check 3: does the bridge balance depend on the emf?

Now check three, the Wheatstone bridge. In the figure, four resistors, P, Q, R and S, form a diamond, with corners A, B, C and D marked by dots. The cell, capital E, is joined across A and C, and the galvanometer, capital G in a circle, sits in the middle arm, between B and D, okay? The bridge is balanced when I sub G, the galvanometer current, with a small G below, is zero. That happens when V sub B equals V sub D: the points B and D are at the same potential. Right? And the condition for that is P over Q equals R over S, as the figure also shows on the right. The figure also notes that at balance the arm B D may be removed, or shorted. Clear? Now the check. Replace the cell with one of a larger emf. Does the bridge stay balanced? It stays balanced, exactly! There is no E in the condition. The balance depends only on the ratio of the arm resistances, not on the emf. Got it?

3:21Check 4: why a potentiometer for emf?

Now check four, the potentiometer. At the top of the figure, the driver cell, epsilon naught, the Greek epsilon with a small zero, sends a steady current through a rheostat and the uniform wire, from A to B, okay? Below, the cell we test, of emf capital E, and the galvanometer run from A to the jockey, J, a sliding contact. With me? Here is the check. To measure the emf of a cell, why is a potentiometer preferred to a voltmeter? First, the voltmeter. It draws a small current, I, from the cell, so it reads the terminal voltage: V equals E minus I times r, where small r is the internal resistance. Okay? Now slide the jockey till the galvanometer reads zero. That is the null point, and the length up to it, small l, is the balancing length. At the null point, no current is drawn from the cell. Right? So I is zero, and we get the emf itself, not the terminal voltage. Then E equals k times l, where k, the potential gradient, is V sub A B, the potential difference across the whole wire, over capital L, the length of the wire, as on the right of the figure. Clear?

4:37Example 1: drift speed at fixed current vs cross-section

Now, Example one, the easier one. At a fixed current, does the drift speed depend on the cross-section? Pause the video and try it first. Done? Okay, let's check. You may remember that the drift speed does not depend on the area of cross-section. So the answer is no, correct? Oh no, careful, that result needs a fixed voltage. With a potential difference V across a length l, v d, the v with a small d, equals e times tau times V, over m times l, with no area in it. Here e is the electron's charge, m its mass, and tau, the Greek letter, the relaxation time. Okay? But here the current is fixed, so use I equals n times e times capital A times v d, as in the figure, where the electron, e minus, drifts slowly against the field, E with an arrow on top. Here n is the number of free electrons per unit volume, and capital A the area of cross-section; the other formulas in the figure are not needed today. The double arrow means which gives: v d equals I, over n times e times capital A, right? Same metal and same current, so n, e and I are fixed, and v d is proportional to one over capital A; the sign on screen means proportional to. With me? So the answer is yes: the drift speed is larger where the wire is thinner. Always ask first: is the current fixed, or the voltage? Very good!

6:16Example 2: does heat in charging a capacitor depend on R

Now Example two, a little harder, and it is our puzzle from the start. Does the heat produced in charging a capacitor fully from a cell depend on the resistance? Pause the video and try it first. Done? Okay, let's check. The heat in a resistor is I squared R, where the small two is a power. So a bigger R gives more heat, correct? No, no, wait, the current is not fixed: a bigger R gives a smaller current, flowing for longer. Look at the left graph, charge q against time t: q equals C E, times, open bracket, one minus e to the power minus t over tau, close bracket. Here C is the capacitance, E the emf of the cell, and small e the exponential base. It settles at C E, the dashed line, and after one time constant, tau, equal to R times C, it reaches zero point six three of C E, okay? The right graph shows discharging, not needed here. So think in energy instead. The cell drives the full charge, C E, through its emf, so it supplies C E times E, which is C E squared. Right? But the capacitor stores only one half C E squared. With me? So the heat in R is C E squared, minus one half C E squared, which gives one half C E squared. Look, there is no R in it! Half the energy from the cell is always lost as heat, whatever R is. Clear? So the answer is no. Only the time taken depends on R, through tau equals R C: a bigger resistor charges it more slowly, with the same heat. And that solves our puzzle. Very good!

8:05Quick revision: Ohm's law, manganin, bridge, potentiometer

So, quick revision. Ohm's law holds only for ohmic conductors, and manganin has a tiny alpha and a high rho. The bridge balances when P over Q equals R over S, whatever the emf, and a potentiometer draws no current at balance, so it reads the emf. At a fixed current, v d equals I, over n times e times capital A, and the heat in charging a capacitor is one half C E squared, whatever R is. Now try the practice question on this page. All the best!