AC voltage across a resistor

A voltage that keeps reversing

Narrated video. Transcript:

The socket in your wall does not give a steady voltage like a cell. Its voltage rises, falls, reverses and comes back, fifty times every second. This is alternating voltage, or ac.

Most of the world’s electricity is ac. One big reason: ac voltages are easy to step up and down with transformers, so power can be sent over long distances cheaply.

Plot the source voltage against time. It is a sine curve: v equals v m sin ωt. Here v m is the peak voltage, the amplitude, and ω is the angular frequency, 2π times ν.

In India the mains frequency ν is 50 hertz. So ω is 2π × 50, about 314 radians per second, and one cycle takes 20 milliseconds.

Now connect this source across a resistor R. At every instant, the loop rule gives v m sin ωt equals i R.

So the current is i equals v m over R, times sin ωt. Write it as i equals i m sin ωt, where the current amplitude i m equals v m over R. Ohm’s law works for ac too.

Draw the current curve on the same graph. It rises when the voltage rises, peaks at the same moment, and passes through zero at the same moment. In a pure resistor, voltage and current are in phase.

Watch the current over one full cycle. For half the cycle it flows one way, for the other half the same amount flows back. So the average current over a cycle is zero.

But the resistor still gets hot! Heating does not care about the direction of the current. So what is the average power? That is the next part.

Average power and rms values

Narrated video. Transcript:

The power dissipated in the resistor at any instant is p equals i²R. With i equals i m sin ωt, that is p equals i m² R sin² ωt.

Plot p against time. Since sin² ωt is never negative, the power is never negative. It swings between zero and i m² R, twice every cycle.

What is the average of sin² ωt over a cycle? Draw the line at one half. The bumps above the line exactly fill the dips below it. So the average of sin² ωt is ½.

So the average power is p bar equals ½ i m² R. We want to write this just like the dc formula, P equals I²R.

So we define the root mean square, or rms, current: I equals i m over √2, which is 0.707 i m. Then the average power is simply P equals I²R.

The rms current is the steady dc current that would produce the same average heating in the resistor. In the same way, the rms voltage is V equals v m over √2, or 0.707 v m.

Then V equals I R and P equals V I, exactly as for dc. When we say the mains supply is 220 volts, we mean its rms value. Its peak is √2 × 220, about 311 volts.

Example: a light bulb is rated 100 watts for a 220 volt supply. Its resistance is R equals V² over P, that is 220² over 100, which is 484 ohms.

The peak voltage of the source is 1.414 × 220, about 311 volts. And the rms current is I equals P over V, 100 over 220, about 0.45 amperes.

From now on, unless we say otherwise, ac voltages and currents mean rms values. Meters on ac also read rms values.

Try it: ac through a resistor

Interactive simulation. Things to try:

• This is the 100 W, 220 V bulb: R = 484 Ω. Watch v (amber) and i (blue) peak at exactly the same moment.
• Look at the power graph: p = vi is never negative. The dashed line, the average, sits at exactly half the peak power.
• Check the readouts: I = 455 mA (0.455 A) and P = VI = 100 W. Now set R = 48.4 Ω (a 1000 W heater).
• Change the frequency. For a resistor, does the current or the phase change at all?

Peak of the mains

Household mains in India is 220 V. What is the peak value v_m of this voltage?

1. 156 V
2. 220 V
3. 311 V
4. 440 V

A room heater

A 1000 W room heater works on a 220 V ac supply. What are the rms current and the peak current through it?

1. I = 4.55 A, i_m = 6.43 A
2. I = 6.43 A, i_m = 4.55 A
3. I = 4.55 A, i_m = 3.21 A
4. I = 4.55 A, i_m = 4.55 A

Summary

• Source and current (resistor): v = v_m sin ωt, i = i_m sin ωt, i_m = v_m/R
• Average power: P = ½ i_m² R = I²R = VI
• rms values: I = i_m/√2 = 0.707 i_m, V = v_m/√2 = 0.707 v_m

• An ac voltage varies like a sine function of time. Indian mains: 220 V (rms), 50 Hz, so the peak is about 311 V.
• In a resistor, current and voltage are in phase, and Ohm’s law holds at every instant.
• The average current over a cycle is zero, but the average power is not: p = i²R is never negative.
• The rms value is the dc value that gives the same average heating. With rms values, P = VI = I²R just as for dc.