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# Innova Junior College

## Tutorial 13 Alternating Current

Learning Outcomes
(a) show an understanding and use the terms period, frequency, peak value and
root-mean-square value as applied to an alternating current or voltage
(b) deduce that the mean power in a resistive load is half the maximum power
for a sinusoidal alternating current.
(c)

## represent an alternating current or an alternating voltage by an equation

of the form x = x0sint.

(d) distinguish between r.m.s. and peak values and recall and solve problems using
the relationship
Irms = I0 / 2 for the sinusoidal case.
(e) show an understanding of the principle of operation of a simple iron-cored
transformer and recall and solve problems using Ns/Np = Vs/Vp = Ip/Is for an
ideal transformer.
(f)

## explain the use of a single diode for the half-wave rectification of an

alternating current.

## Part I: Characteristics of Alternating Currents

1. N04/I/26
L/O: (b) Deduce that the mean power in a resistive load is half the maximum power
for a sinusoidal alternating current.

## There is a sinusoidal alternating current in a resistor. What is the mean

power dissipated in the resistor?
A

1
2
( I 0)
2

1
Po
2

1
2
(I )
2 0

1
P
2 0
2. N06/I/25
L/O: (c) represent an alternating current or an alternating voltage by an equation
of the form x = x0sint.

## A cathode-ray oscilloscope screen with a grid of 1 cm squares displays an

alternating voltage waveform. The settings of the oscilloscope are: gain =
0.50 V cm-1, time-base = 5.0 ms cm-1.

A

V = 1.0 cos 50 t

## V = 1.0 cos 130 t

V = 1.0 cos 630 t

## V = 1.0 cos 310 t

3. N07/I/32
L/O: (d) distinguish between r.m.s. and peak values and recall and solve problems
using the relationship
Irms = I0 / 2 for the sinusoidal case.

## A sinusoidal alternating current has a peak value of I0. Which expression is

correct for the root-mean-square current Irms?
A

I rms =

I0
2

I rms =

I0
2

I rms = 2 I 0

I rms =2 I 0

4. N08/I/34
L/O: (d) distinguish between r.m.s. and peak values and recall and solve problems
using the relationship
Irms = I0 / 2 for the sinusoidal case.

## The peak value of the current is I0.

What is the root-mean-square value?

5. N09/I/34

## L/O: (c) represent an alternating current or an alternating voltage by an equation

of the form x = x0sint.

The graph shows how an alternating current I of peak value I0 varies with
time t.

## Which expression gives the alternating current I?

A

I =I 0 sin ( 5 t )

I =I 0 sin ( 800 t )

I =I 0 sin (

2 t
)
2.5

I =I 0 sin (

t
)
0.0025

6. N09/I/35
L/O: (b) deduce that the mean power in a resistive load is half the maximum power
for a sinusoidal alternating current.

## A sinusoidally-alternating voltage supply at 50 Hz connected across a load

resistor of 200 delivers a peak current of 2.0 A. The frequency of the
supply is doubled to 100 Hz.
What is the mean power dissipated in the load resistor at the higher
frequency?
A

200 W

400 W

800 W

1600W

7. N10/I/33
L/O: (a) show an understanding and use the terms period, frequency, peak value and
root-mean-square value as applied to an alternating current or voltage, & (d)
distinguish between r.m.s. and peak values and recall and solve problems using the
relationship Irms = I0 / 2 for the sinusoidal case.

The graph shows how the potential difference across an alternating supply
varies with time.

## What are the frequency f and the root-mean-square potential difference

Vrms of the a.c. supply?
f / Hz

Vrms / V

40

156

40

78

400

156

400

78

8. N11/II/3
L/O: (a) show an understanding and use the terms period, frequency, peak value and
root-mean-square value as applied to an alternating current or voltage, & (d)
distinguish between r.m.s. and peak values and recall and solve problems using the
relationship Irms = I0 / 2 for the sinusoidal case.

## A cathode ray oscilloscope (c.r.o.) is used to show the potential difference

(p.d.) across R. The screen of the c.r.o. displays the variation with time of
the p.d. across R, as shown in Fig. 3.2.

On the vertical scale, 1.0 cm represents 5.0 V. On the horizontal scale, 1.0
cm
represents 10 ms.
(a) Use Fig. 3.2 to determine
(i)
the frequency of the a.c. supply,
(ii)

Calculate
(i)

(ii)

## Part II: Transformers

9. N02/I/27
L/O: (e) show an understanding of the principle of operation of a simple iron-cored
transformer and recall and solve problems using Ns/Np = Vs/Vp = Ip/Is for an ideal
transformer.

## 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.

6.0

A 240 V a.c. supply is connected to the primary coil and a 6.0 resistor is
connected to the secondary coil.
A

0.10 A

0.14 A

2.0 A

40 A
10.

N03/I/24

## L/O: (e) show an understanding of the principle of operation of a simple iron-cored

transformer and recall and solve problems using Ns/Np = Vs/Vp = Ip/Is for an ideal
transformer.

## The primary coil of a transformer is connected to an alternating voltage

supply. The secondary coil is connected across a variable resistor.

Which change will cause a decrease in p.d. across the secondary coil?
A increasing the cross-sectional area of the secondary coil
B increasing the current in the primary coil
C increasing the number of turns of the primary coil
D increasing the resistance of the variable resistor
11.

N07/I/33

## L/O: (e) show an understanding of the principle of operation of a simple iron-cored

transformer and recall and solve problems using Ns/Np = Vs/Vp = Ip/Is for an ideal
transformer.

## A 100% efficient transformer is connected as shown to a sinusoidal a.c.

supply.

0.16 A

0.80 A

2.8 A

4.0 A
12.

N02/III/4(part)

L/O: (a) show an understanding and use the terms period, frequency, peak value and
root-mean-square value as applied to an alternating current or voltage, & (e) show
an understanding of the principle of operation of a simple iron-cored transformer
and recall and solve problems using Ns/Np = Vs/Vp = Ip/Is for an ideal transformer.

(a) Explain what is meant by the terms peak voltage and root-meansquare voltage when applied to a sinusoidal alternating voltage.
[3]
(b) The equation V = 564 000 sin 100 t represents a sinusoidal
alternating voltage for an overhead cable on an electrical distribution
system. State the frequency, peak voltage and root-mean-square voltage
for this alternating voltage.
[3]
(c) Explain why such a high voltage is advantageous for the transmission
of electrical energy.
[
3]
(d) Describe briefly how such voltages are obtained in practice.
[3]
(e) A cable used for the transmission detailed in (b) has a circular crosssection of radius 0.012 m. Fig. 8 is a full-scale drawing showing the
electric field surrounding the cable together with lines of equal potential
at an instant when the potential of the
cable is +564 000 V.
(i)

## State the equation relating field strength with potential gradient.

(ii)
Use Fig. 9 to estimate the potential gradient near the surface of the
cable.
(iii)

Explain why a cable of larger radius, but at the same potential, will
have a smaller electric field at its surface.

[5]

Fig. 9

(f) Sketch a diagram suggesting electric field lines near a cable of square
cross-section.
[3]

13.

N04/III/6(part)

L/O: (a) show an understanding and use the terms period, frequency, peak value and
root-mean-square value as applied to an alternating current or voltage, & (e) show
an understanding of the principle of operation of a simple iron-cored transformer
and recall and solve problems using Ns/Np = Vs/Vp = Ip/Is for an ideal transformer.

V r.m.s.
(i)

[1]

(ii)

[2]

(iii)

[1]

## Part III: Rectification with a Single Diode

14.

NJC Prelim/2011/I/31

L/O: (f) explain the use of a single diode for the half-wave rectification of an
alternating current.

For the circuit shown above, which of the following graphs shown below
best represents the variation of current I across the resistor with time t?

15.

NYJC Prelim/2011/I/35

L/O: (f) explain the use of a single diode for the half-wave rectification of an
alternating current.

## In the given circuit, if the sinusoidal a.c. source has a peak-to-peak

voltage of 20 V, what is the r.m.s. current through the 50 resistor?

0.07 A
0.20 A

0.10 A

0.14 A
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