Overview

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. Take a five ohm wire, and stretch it to three times its length. Three times the length, so three times the resistance, fifteen ohm, right? Hmm, hold that thought; Example one will settle it. This chapter has three sets of mixed problems in the JEE Main style, so let us see how to use them. Let's begin.

0:27How this chapter works

So, why this chapter? Current electricity usually gives two or three questions in a JEE Main shift, of a few familiar types. Okay? First, a resistivity or drift calculation. Then an equivalent resistance, or a cell or Kirchhoff problem. Right? Or a meter, bridge or potentiometer question, and often a power or R C item, with a resistor and a capacitor. Now let me ask you this: our stretched wire, which type is it? Yes, exactly, it is the first type on the list, a resistivity question! Now the sets: Set one is current and networks, Set two is cells and instruments, and Set three is power and R C circuits. Clear? Real JEE Main questions, from twenty twenty one to twenty twenty six, are at the end of Chapters one to six. Here, each set has ten original solved problems, one to two minutes each. With me?

1:25Set 1 tools: current, resistivity and networks

Now, the tools for Set one. The current, I, equals n times e times capital A times v d. Here n is the number of free electrons per unit volume, e the electron's charge, capital A the area of cross-section, and v d, the v with a small d below, the drift speed. Okay? Ohm's law, V equals I times R, holds for ohmic conductors at constant temperature; V is the potential difference, and R the resistance. Right? Then R equals rho l over capital A, with l the length. Rho, the Greek letter like a curly p, is the resistivity: it depends only on the material and the temperature, but R also depends on the length and the area. Clear? For networks: in series, resistances add, R one plus R two; the small one and two label the resistors. In parallel, it is R one times R two, over R one plus R two, always less than the smaller resistance. Got it?

2:23Set 2 tools: cells, Kirchhoff and instruments

Now, Set two. Take a real cell of e m f capital E and internal resistance small r, driving an outside resistor R; the current is E over R plus r. The terminal voltage V is E minus I r while the cell discharges, and E plus I r while it is being charged. Okay? Kirchhoff's junction rule is charge conservation: the current in equals the current out. The loop rule is energy conservation: round any closed loop, the potential changes add up to zero. Right? Use one sign convention throughout: across a resistor along the current, write minus I R; through a cell from its minus to its plus terminal, write plus E, clear? Now the meters. For an ammeter, a galvanometer gets a small shunt in parallel; for a voltmeter, a large resistance R in series. So tell me: to measure a cell's e m f, which is better, a voltmeter or a potentiometer? The potentiometer, very good! At balance it draws no current, so it measures the e m f itself: E equals k times l, with k the potential gradient and l the balancing length. But a voltmeter draws some current, so it reads the terminal voltage instead. And a Wheatstone bridge balances when P over Q equals R over S, for its four arms. Got it?

3:47Set 3 tools: power and RC circuits

Now, Set three. Power P equals V times I, or I squared R, or V squared over R; the small raised two means squared. And one kilowatt hour, the k W h, is three point six times ten to the power six joules, with capital J for joule. With me? Use I squared R when the current is common, as in series, and V squared over R when the voltage is common, as in parallel. Now, two bulbs rated for the same voltage but different powers are joined in series. Which one glows brighter? Oh no, careful, it is not the higher-rated one! The lower power rating means the larger resistance. In series the current is the same, so the power goes with R, and the lower-rated bulb glows brighter. Right? Next, in a D C circuit in the steady state, no current flows into a capacitor, so its branch is an open circuit. Clear? While it charges through R from a cell of e m f E, q equals C E, times one minus e to the power minus t over tau. Here q is the charge, C the capacitance, t the time, and small e the base of natural logarithms. Tau, the Greek letter, is R times C, the time constant. Okay?

5:00Example 1: 5 ohm wire stretched to 3 times its length

Now, Example one, an easy one from Set one, and it is our opening puzzle. Here it is: a five ohm wire is stretched to three times its length. Find its new resistance. The omega sign means ohm. Pause the video and try it first. Done? Okay, let's check. Three times the length, so three times the resistance, fifteen ohm. Is that right? No, no, wait, that is the trap! Stretching adds no metal, so the volume stays fixed. The length becomes three l, so the area becomes capital A over three; the arrow means becomes. Okay? Now use rho l over capital A again. R prime, the R with a small dash, is rho times three l, over capital A over three. One factor of three comes from the length and another from the area, so we get nine times R, right? So the new resistance is nine times five, which gives forty five ohm, not fifteen. So R is proportional to l squared; the sign like an open Greek alpha means proportional to. Three squared is nine, and our puzzle is solved. Clear?

6:12Example 2: time for a capacitor to reach 75% charge

Now Example two, a medium one from Set three. R is two megaohm, capital M with the omega, and C is five microfarad, the Greek mu with capital F for farad. How long does the capacitor take to reach seventy five percent of its final charge? Pause the video and try it first. Done? Okay, let's check. Tau equals R times C: two times ten to the power six, times five times ten to the power minus six, which gives ten seconds, with s for seconds. With me? Now, seventy five percent of the charge, so seventy five percent of tau, seven point five seconds. Is that right? No, it is not, and the charging curve, the left graph, shows why; the right graph is for discharging. The charge does not grow in a straight line: after one tau, ten seconds, it is only at zero point six three C E, the dashed level. So we need more than ten seconds: set one minus e to the power minus t over tau equal to zero point seven five; then e to the power minus t over tau is one quarter. Okay? Take the natural log, written l n: t over tau is the natural log of four. So t is tau times natural log four, ten times one point three eight six, which gives about thirteen point nine seconds; the wavy equals sign means approximately. That is more than one tau, as the graph said. Got it?

7:35Quick revision: R proportional to l squared, 63% at RC

So, quick revision. First, spot the type, then use that set's tool. Second, stretching keeps the volume, so R is proportional to l squared, and a potentiometer draws no current at balance. Third, a capacitor is an open circuit in the steady state, and after one time constant, R times C, it is sixty three percent charged. Now open Set one, give each problem one to two minutes, and try each set's practice problem. All the best!