Transformers and Power Transmission

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Power lines into a town can carry eleven thousand volts, but the sockets in your home give only two hundred and twenty. So why not send two hundred and twenty volts all the way? By the end of this video, you will see how ten times the voltage cuts the power wasted in the wires a hundred times. The secret is the transformer, and it comes up often in JEE, so let's begin.

0:28How a transformer works

Here is a transformer. The circle with a wave is an alternating source, driving the dark blue primary coil with N p turns; that is capital N with a small p below. The orange secondary coil has N s turns and feeds the load. Both sit on one iron core, and no wire joins them. Okay? So how does energy get across? The alternating current in the primary sets up a changing flux in the core, which induces an emf in the secondary. That is mutual induction, so a transformer works only with alternating current, the A C on screen. Right? Now tell me this. What happens if we connect a battery, a steady D C voltage, to the primary? Once the current is steady, nothing appears at the secondary, that is correct! Steady current means constant flux, and a constant flux induces no emf. Clear? Now the main formula on screen: V s over V p equals N s over N p, the turns ratio. V p and V s are the primary and secondary voltages; the small letters below name the coil. So the voltages go in the ratio of the turns. With me? More turns on the secondary, N s greater than N p, means the voltage goes up: a step-up transformer. N s less than N p gives a step-down transformer. In the figure, the orange coil has more turns, so this one steps up.

1:54Ideal transformer: no power is gained

Got it? Here is a question that catches many students. If a step-up transformer raises the voltage ten times, does it give ten times the power too? Oh no, it cannot, because a transformer does not create energy. In an ideal transformer, power in equals power out: V p times I p equals V s times I s, as the figure writes at the top. I p and I s are the primary and secondary currents, okay? Rearrange it, and I s over I p equals N p over N s, the turns ratio upside down. So the current does the opposite of the voltage: ten times the voltage means one tenth the current, and no extra power. Right? One more result worth knowing. Put a load R on the secondary; seen from the primary, it looks like R times N p over N s, the whole bracket squared. So with N p to N s of ten to one, the load looks one hundred times bigger. Clear?

2:54Real transformers: efficiency and losses

But real transformers are never perfect. The efficiency, eta, the Greek letter like an n with a long tail, is the output power, V s times I s, over the input power, V p times I p. Okay? Some power is always lost, so where does it go? First, copper loss: the current heats the windings, I squared R in the wires, and thick wire keeps it small. With me? Now look at the figure: it says laminated iron core. Laminated means made of thin sheets. Why not use one solid block of iron? Because of eddy currents, yes, exactly! The changing flux also induces currents swirling inside the iron, wasting energy as heat. Thin sheets cut these currents down. Got it? Next, hysteresis loss: the core is magnetised one way and then the other, every cycle, and that costs energy. Soft iron reduces it. Right? And last, flux leakage: some flux from the primary misses the secondary, and winding the coils over each other reduces it. Four losses, four fixes; remember this list.

4:04Why power is sent at high voltage

Now, back to our puzzle. For a fixed power P sent at a voltage V, the line current, I, is P over V, okay? The wires have a resistance R, so the power lost in the line is I squared R, right? So, to waste less power, should we send it at a high voltage or a low one? Yes, exactly, we send it at a high voltage! The double arrow you see on screen means implies. For the same power, high voltage gives a small current, and the loss goes as the current squared. So ten times the voltage means one hundredth of the loss. Clear? That is where transformers come in. At the power station, a step-up transformer raises the voltage for the long journey; in the town, a step-down transformer brings it down to two hundred and twenty volts. All this needs alternating current. With me?

4:56Example 1: 2200 V to 220 V, secondary turns, primary current

Now, our first example. This transformer steps two thousand two hundred volts down to two hundred and twenty volts; the capital V means volts. The primary has two thousand turns. Find the secondary turns. And if the secondary supplies ten amperes, the capital A, find the primary current of this ideal transformer. Pause the video and try it first. Are you done? Okay, let's check. N s equals N p times V s over V p, so two thousand times two hundred and twenty over two thousand two hundred. The voltage falls to one tenth, so the turns do too, giving two hundred turns. Okay? Now the primary current. The voltage drops ten times, so many students do ten times ten and get one hundred amperes. Is that right? No, no, careful, that is the trap! Power in equals power out, so the currents go the opposite way to the voltages. I p over I s equals V s over V p, the voltage ratio flipped. With me? So I p equals ten times two hundred and twenty over two thousand two hundred, which gives one ampere. As a check, two thousand two hundred times one, and two hundred and twenty times ten, both give two thousand two hundred watts. Got it?

6:16Example 2: line loss of 100 kW sent at 1000 V and 10000 V

Now, remember our puzzle? Here it is in numbers. One hundred kilowatts, the k W on screen, goes along a line of total resistance five ohms, the capital omega. Find the power lost when it is sent at one thousand volts, and at ten thousand volts. Pause the video and try it first. Are you done? Okay, part a first. Some students write the loss as V squared over R, so one thousand squared over five, which is two hundred kilowatts. Is that right? No, it cannot be: two hundred kilowatts lost, when we sent only one hundred! The one thousand volts is the sending voltage, not the voltage across the line. So find the current: I equals P over V, one hundred thousand over one thousand, which gives one hundred amperes. Okay? Now the loss is I squared R: one hundred squared, times five, which is fifty thousand watts; the capital W means watts. That is fifty kilowatts, fifty percent, half the power gone as heat! With me? Now part b, at ten thousand volts. I equals one hundred thousand over ten thousand, which gives ten amperes. Right? The loss is ten squared, times five, which is five hundred watts, or zero point five kilowatts, just zero point five percent! So, mystery solved: ten times the voltage, and a hundred times less loss. Clear?

7:40Quick revision: turns ratio, VpIp = VsIs, high voltage lines

So, quick revision. V s over V p equals N s over N p, and a transformer works only with alternating current. Ideally, power in equals power out, so voltage up means current down. And sending power at high voltage keeps the current, and the I squared R loss, small. Now go and try the practice problems on this page. All the best!