Page 468 - Physics Coursebook 2015 (A level)

Cambridge International A Level Physics
456
WORKED EXAMPLE
QUESTIONS
7 What is the r.m.s. value of an alternating current represented (in amps) by the equation
I = 2.5 sin (100πt)?
8 The mains supply to domestic consumers in many European countries has an r.m.s. voltage Vrms
of 230 V. (Note that it is the r.m.s. value which is generally quoted, not the peak value.) What is the peak value of the supply?
Calculating power
The importance of r.m.s. values is that they allow us
to apply equations from our study of direct current to situations where the current is alternating. So, to calculate the average power dissipated in a resistor, we can use the usual formulae for power:
P = I2R = IV = V2 R
Remember that it is essential to use the r.m.s. values of I and V, as in Worked example 1. If you use peak values, your answer will be too great by a factor of 2.
Where does this factor of 2 come from? Recall that r.m.s. and peak values are related by I0 = 2Irms. So, if you calculate I 2R using I0 instead of Irms, you will introduce a factor of 22, i.e. a factor of 2. (The same is true if you calculate power using V0 instead of Vrms.) It follows that, for a sinusoidal alternating current, peak power is twice average power.
1 A sinusoidal p.d. of peak value 25 V is connected across a 20 Ω resistor. What is the average power dissipated in the resistor?
Step1 Calculatether.m.s.valueofthep.d.: V r m s = V 02 = 2 52 = 1 7 . 7 V
QUESTIONS
9 What is the average power dissipated when a sinusoidal alternating current with a peak value of 3.0 A flows through a 100 Ω resistor?
10 A sinusoidal voltage of peak value 325 V is connected across a 1 kΩ resistor.
a What is the r.m.s. value of this voltage?
b Use V = IR to calculate the r.m.s. current which
flows through the resistor.
c What is the average power dissipated in the resistor?
d What is the peak power dissipated in the resistor?
Explaining root-mean-square
We will now briefly consider the origin of the term root-mean-square and show how the factor of 2 comes about. The equation P = I2R tells us that the power P is proportional to the square of the current I. Figure 29.10 shows how we can calculate I2 for an alternating current. The current I varies sinusoidally, and during half of each cycle it is negative. However, I2 is always positive (because the square of a negative number is positive). Notice that I2 varies up and down, and that it has twice the frequency of the current.
Now, if we consider <I2>, the average (mean) value of I2, we find that its value is half the peak value (because the graph is symmetrical):
< I 2 > = 12 I 2 I
0t
I2 I2
<I 2>
0
Step 2 Now calculate the average power dissipated: P=V2=17.72=15.6W
R 20
(Note that, if we had used V0 rather than Vrms, we
would have found P = 252 = 31.3 W, which is double
t
20 the correct answer.)
Figure 29.10 An alternating current I is alternately positive and negative, while I 2 is always positive.