CAIE A-Level Physics A2 21.1 Characteristics of Alternating Currents Questions

Practise analysing sinusoidal current and voltage, period, frequency, peak and rms values, and calculating mean or instantaneous resistive power.

Exam points

analyse period, frequency, peak and rms values and use sinusoidal current or voltage equations
calculate mean and instantaneous power in a resistive load

Covered syllabus points

21.1.1—The terms period, frequency and peak value as applied to an alternating 21.1.2—Equations of the form x = x0 sin ωt representing a sinusoidally alternating 21.1.3—The fact that the mean power in a resistive load is half the maximum power 21.1.4—Root-mean-square (r.m.s.) and peak values and recall and use I r.m.s. = I0 /

[Maximum number: 4]

(a)

The output of a power supply is represented by:

V=9.0 \sin 20 t

where V is the potential difference in volts and t is the time in seconds.
Determine, for the output of the supply:

[ 3 ]

(i)

the root-mean-square (r.m.s.) voltage, $V_{\text {r.m.s }}$ .

V_{\text {r.m.s }}=

[ 1 ]

$9.0 / \sqrt{ } 2=$
6.4 V

(ii)

the period T.

[ 2 ]

$\begin{aligned} \omega=20 \omega=2 \pi / T T=2 \pi / 20 \end{aligned}$ $T=0.31 \mathrm{~s}$

(b)

The variations with time t of the output potential difference V from two different power supplies are shown in Fig. 9.1 and Fig. 9.2.

Fig. 9.1

Fig. 9.2

The graphs are drawn to the same scale.
State and explain whether the same power would be dissipated in a $1.0 \Omega$ resistor connected to each power supply.

[ 1 ]

[Maximum number: 6]

(a)

Two coils of insulated wire are wound on an iron bar, as shown in Fig. 8.1.

Fig. 8.1

There is a current $I_{1}$ in coil 1 that varies with time t as shown in Fig. 8.2.

Fig. 8.2

[ 6 ]

(i)

The variation with t of $I_{1}$ can be represented by the equation

I_{1}=X \sin Y t

where X and Y are constants.

Use Fig. 8.2 to determine the values of X and Y. Give units with your answers.

\begin{aligned} & X=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . . . . . . . . . . . . . . . . . u n i t \\ & Y=\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . . . . . . . . . . . . . . . u n i t \end{aligned}

[ 3 ]

$X=0.85 \mathrm{~A}$ $Y=2 \pi / 0.040$ $=160 \mathrm{rad} \mathrm{s}^{-1}$

(ii)

The current in coil 1 gives rise to a magnetic field in the iron bar.

Assume that the flux density of this magnetic field is proportional to $I_{1}$ .
An alternating electromotive force (e.m.f.) is induced across coil 2. The p.d. across coil 2 is measured using the voltmeter and has a root-mean-square (r.m.s.) value of 4.6 V .

On Fig. 8.3, sketch a line to show the variation with t of $V_{2}$ between t=0 and $t=0.08 \mathrm{~s}$ .

Fig. 8.3

[ 3 ]

two cycles of a sinusoidal curve with a period of 0.040 s
correct phase (i.e. $V_{2} \max / \min$ at t=0,0.02,0.04,0.06 and 0.08 s , and $V_{2}$ zero at $t=0.01,0.03,0.05,0.07 \mathrm{~s}$ )
maximum / minimum $V_{2}$ shown (consistently) at $\pm 6.5 \mathrm{~V}$

[Maximum number: 5]

An ideal transformer is shown in Fig.9.1.

Fig. 9.1

(a)

A resistor is connected between the output terminals of the secondary coil. The mean power dissipated in the resistor is 1.2 W . It may be assumed that the varying voltage across the resistor is equal to the varying e.m.f. E shown in Fig. 9.2.

[ 5 ]

(i)

Calculate the resistance of the resistor.

resistance =

$\Omega$

[ 2 ]

either

mean power $=V^{2} / 2 R$ and $V=52(\mathrm{~V})$ $\begin{aligned} R =52^{2} /(2 \times 1.2) =1100(1127) \Omega \end{aligned}$
or

mean power $=V^{2} / R$ and $V=52 / \sqrt{2}(=36.8 \mathrm{~V})$ $\begin{aligned} R =36.8^{2} / 1.2 =1100 \Omega \end{aligned}$

(ii)

On Fig.9.3, sketch the variation with time t of the power P dissipated in the resistor for t=0 to $t=22.5 \mathrm{~ms}$ .

Fig. 9.3

[ 3 ]

[Maximum number: 2]

The output of a power supply is represented by:

V=9.0 \sin 20 t

where V is the potential difference in volts and t is the time in seconds.
Determine, for the output of the supply:

the period T.

$\begin{aligned} \omega=20 \omega=2 \pi / T T=2 \pi / 20 \end{aligned}$ $T=0.31 \mathrm{~s}$

[Maximum number: 1]

An electron in a metal rod moves randomly about a mean position. When an alternating voltage is applied to the ends of the rod, the mean position can be considered to oscillate with simple harmonic motion along the axis of the rod. Fig. 4.1 shows the variation with time t of the displacement x of the mean position from a fixed point on the axis of the rod.

Fig. 4.1

The rod has a cross-sectional area of $4.3 \mathrm{~cm}^{2}$ and contains a number density of conduction electrons (charge carriers) of $8.5 \times 10^{28} \mathrm{~m}^{-3}$ .

All of the conduction electrons in the rod may be assumed to be oscillating in phase with, and with the same amplitude as, the oscillation shown in Fig. 4.1.

Determine the root-mean-square (r.m.s.) current in the rod.
r.m.s. current = A

$\begin{aligned} I_{\text {r.m.s. }} =I_{0} / \sqrt{ } 2 =0.64 / \sqrt{ } 2 =0.45 \mathrm{~A} \end{aligned}$

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Covered syllabus points

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