AC Tutorial

Physics JC1 2009
Tutorial 12: Alternating Currents
Period, frequency, peak value and root-mean-square value
1) N97/2/6 (a) (b)
An alternating current varies with time in the way shown in figure below.
Use the graph to determine, for this alternating current,
(i) frequency
(ii) the peak value
(iii) the root-mean-square value
Copy the figure above into your answer sheet. On the same plot, sketch the graph which
shows how the power supplied by this current to a resistor of resistance 5  varies with
time. Label the vertical axis as power and mark on this axis the maximum value of the
The current is a sinusoidal function.
(i) From the graph,
T = 20 ms
Frequency f 
= 50 Hz
20  103
(ii) Peak current, I0 = 3.0 A
(iii) Irms 
 2.12 A
I/ A
time / ms
time / ms
P0 = I02R = 32 x 5 = 45 W
2) J88/1/23
When an a.c. supply of 240 V r.m.s. is connected to the terminals PQ in the circuit shown
below, the fuse F breaks the circuit if the current just exceeds 13 A r.m.s.
When the a.c. supply is replaced with a 120 V d.c. source, an identical fuse breaks the
circuit if the current just exceeds
B. 13/ 2 A
A. 13/2 A
C. 13 A
D. 13
2 A
E. 26 A
Explain why you make your choice.
Solution: C
Fuse breaks due to the heating effect. Since the heating effect produced by the a.c. and
d.c. is the same, by definition, the r.m.s. current of the ac must equal to that of the d.c.
that produces the same heating effect for identical fuse. The results is independent of the
voltage of the source used (d.c. or a.c.).
3) N00/1/21
A mains electricity supply has root-mean-square voltage of 240 V and a peak voltage of
340 V. When connected to this supply, a heater dissipates energy at a rate of 1000 W.
The heater is then connected to a 340 V d.c. supply and its resistance remains the same.
At what rate does the heater now dissipate energy?
Given that:
Vrms = 240 V
Vpeak = 340 V
Pave = 1000 W
Since the resistance remains the same whether the source is a.c. or d.c., it will be easier
to compute the value in terms of R.
V 2
Pave  rms
1000 
R = 57.6 
With a 340 d.c.,
V 2
P  d .c .
= 2010 W
4) J77/2/23
The figure on the right is the waveform of an
alternative sinusoidal voltage of amplitude 200 V.
The r.m.s value of the voltage is
A. 0 V
D. 141.4 V
B. 70.7 V
E. 200 V
C. 100 V
Solution: C
For a normal a.c. voltage:
For a half- rectified a.c. voltage:
V2 / V2
V2 / V2
Vr .m.s. 
area of V 2  t graph over 1 period
For a half-rectified a.c. voltage, the area under a V2 – t graph is half of that of a normal
a.c. voltage.
½  area of a full sine graph
Therefore, Vr .m.s. 
area of a full sine graph
 ½
1 V0 V0
Vrms =
2 2
= 100 V
Mean Power of a sinusoidal alternating current
5) Deduce that the mean power in a resistive load is half the maximum power for a
sinusoidal alternating current
Maximum power: Po = Vo Io
Mean power:
<P(t)>  Irms Vrms
Changing r.m.s. variable to peak variable:
 I  V  I V
<P(t)>   o  o   o o  P0
 2  2  2
6) J80/2/25
An alternating current, I = I0 sin ωt, passes through a resistor of resistance R. Which one
of the following best represents the variation with time t of the power P dissipated in the
Solution: C
To obtain P(t) from I(t) graph, make use of the equation P = I2R.
Thus a P(t) graph will follow a I2(t) graph since R is a constant value (with time). Having a
sinusoidal function for I(t), the P(t) should follow a square of a sinusoidal graph.
7) N97/1/21
The diagram shows an iron-cored transformer assumed to be 100 % efficient. The ratio
of the secondary turns to the primary turns is 1:20.
A 240 V a.c. supply is connected to the primary coil and a 6.0  resistor is connected to
the secondary coil.
What is the current in the primary coil?
Ns Vs
Np Vp
 s
20 240
VS = 12 V
Power input to transformer = Power output from transformer
IP  240 
IP = 0.10 A
8) The turns ratio Noutput / Ninput for a particular transformer used in a CD player is 1:13.
When used with 120-V (rms) household services, the transformer draws an r.m.s. current
of 250 mA. Find the
(a) r.m.s. output voltage of the transformer and
(b) power delivered to the CD player.
(a) s  s
Np Vp
13 Vs
1 120
Vs = 9.23 V
(b) Power output from transformer = Power input to transformer
Power output from transformer = IP VP
= 9.23 × 250 × 10-3
9) N91/1/21; J95/1/20
The diagram shows two circuits used to rectify an alternating current. A cathode-ray
oscilloscope (c.r.o.) with fixed settings of time-base and Y-gain is connected in turn as
shown and produces a trace in each case.
Which set of diagrams shows these traces most accurately?
Solution: B
c.r.o. 1 shows a normal a.c. being half rectified.
c.r.o. 2 is a full wave rectifier. Students needs to
trace the flow of current through the diode.
10) N96/2/4
In the circuit shown, the four diodes A, B, C and D are connected to form a bridge
rectifier. The diodes may be assumed to be ideal.
(a) (i) On the figure above, mark with a + the positive output terminal
(ii) State which diodes do not conduct when input terminal X is positive with respect
to input terminal Y.
(a) (i)
Current flows from high potential to low potential. In either of the cycle of the input,
current through the output is from top to bottom.
(ii) Looking at the first diagram, diode B and D do not conduct when terminal X is
positive to terminal Y.
(b) The input terminals X and Y are connected to the secondary coil of an ideal
transformer. The primary coil is connected to a 240 V r.m.s. alternating supply. The
input to the bridge rectifier is to be 12 V r.m.s.
number of turns on secondary coil
(i) What is the value of the ratio
number of turns on primary coil
(ii) An ideal diode does not conduct when a potential difference is connected in one
direction across the diode. The maximum potential difference which can be
applied in this direction across before it conducts is known as the breakdown
voltage of the diode. Calculate the minimum breakdown voltage of the diodes in
the above figure if the circuit is to function correctly?
Ns Vs
(b) (i)
Np Vp
Np 240 20
(ii) Diode must be able to withstand the peak voltage in both directions.
V0  2Vr .m.s.  2  12  17 V
Power Transmission
11) J98/2/5
Electrical power of 4400 kW is supplied to an industrial consumer at a considerable
distance from a generating station. This is shown in the figure below.
In order to do this, the electricity supply company makes use of a circuit containing two
transformers, T and U. The transformers can be considered to be ideal and the supply
cables to have negligible resistance.
(a) The power is generated at 11 kV r.m.s. and is supplied to the consumer at 11 kV
r.m.s. Calculate the r.m.s. current supplied to the consumer.
Power output P = Ir.m.s. V r.m.s.
4 400 000 = Ir.m.s × 11 000
Ir.m.s = 400 A
(b) There is a potential difference of 275 kV r.m.s. between the supply cables. Calculate
(i) the ratio Ns/Np for transformer T and the ratio Ns/Np for transformer U
(ii) the current in the supply cables
(i) Transformer T:
Transformer U:
Ns Vs 275
 25 :1
N p Vp
= 1: 25
N p 25
(ii) For transformer U:
Ns I p
N p Is
25 400
Ip = 16 A
(c) Explain why, when the resistance of the supply cables cannot be neglected, this
arrangement is preferable to a system which generates and transmits the power at
the same voltage of 11 kV r.m.s.
Without the step up and step down transformer, the current in the supply cable will be
approximately equal to 400 A but much larger than 16A. As such, the power
dissipated through the supply wire would be proportional to I2 R. Thus the
corresponding power loss in transmission will be much larger.
Transmission at higher voltage ensures that less power is wasted.
(d) (i) Explain the distinction between the r.m.s.value and the peak value of an
alternating current.
The root-mean-square (rms) value, Irms, of an AC is equivalent to that value of a
steady direct current which would dissipate energy in a given resistance at the same
rate as that provided by the alternating current while peak value is the maximum
value of the alternating current which occurs twice in 1 cycle.
(ii) Find the peak value of the current in (a).
I0  2Ir .m.s  2  400  570 A