2015 Question 6: Purely Inductive & Resistive Circuit

(a) Figure 8 shows the circuit diagram of a coil with an ohmic resistance of 3 ohms and an
inductance of 0.01 henry, connected to a 12 V a.c., 50 hertz supply

Calculate the

i. Inductive reactance of a coil

Xl = 2πfL = 2π × 50 ×0.01H = 3.14Ω
ii. Impedance of the coil
Z = √𝑹𝟐 + 𝑿𝑳𝟐 = √𝟑Ω𝟐 + 𝟑. 𝟏𝟒Ω𝟐 = √𝟗 + 𝟗. 𝟖𝟔 = √𝟏𝟖. 𝟖𝟔 = 4.34Ω
iii. Circuit current
V = I×Z
I = V/Z = 12V/4.34Ω = 2.76A

(b) Briefly explain the relationship between the circuit current and the applied voltage when
the a.c. voltage is applied across a
(i) Purely resistive circuit (1 mark)
In a purely resistive A.C. circuit, the current IR and applied voltage VR are
in phase
(ii) Purely inductive circuit. (1 mark)
In a purely inductive A.C. circuit, the current IL lags the applied voltage VL
by 90◦.
(c) Briefly explain the relationship between the current in the main winding and the starting
winding of a single-phase a.c. motor with a highly
(i) inductive starting winding
In the case of an inductive starting winding, the winding has a higher
inductance compared to the main winding. This causes a phase shift between
the current and voltage applied to the motor. The current in the starting
 winding lags behind the voltage, which creates a rotating magnetic field that
drives the rotor. Once the motor reaches its operating speed, the starting
winding is disconnected from the circuit by a centrifugal switch.
(ii) capacitive starting winding.
In the case of a capacitive starting winding, the winding has a higher
capacitance compared to the main winding. This causes a phase shift
between the current and voltage applied to the motor. The current in the
starting winding leads the voltage, which creates a rotating magnetic field
that drives the rotor. Once the motor reaches its operating speed, the starting
winding is disconnected from the circuit by a centrifugal switch.
2011 Question 6: Inductive Reactance
(a) (i) Define ‘inductive reactance’. (2 marks)
This is the opposition to current flow in a purely inductive circuit.
(ii) Write the formula for inductive reactance and explain EACH of the terms in the
formula. (1 mark)
Formula: XL = 2πFL
XL – inductive reactance
2π – constant
F – frequency
L – inductance
(b) An alternating voltage source is connected across an inductor L as shown in Figure 2.

(i) Use a wave diagram to show the phase relationship between the voltage VL and the
current IL associated with the inductor. (3 marks)
 (ii) Illustrate the relationship between the current IL and voltage VL using a vector
diagram. (7 marks)

(c) An a.c. supply of 110 V and 50 Hz is applied across an inductor of 0.5 H. Calculate the
current IL flowing through the inductor. (5 marks)
XL = 2πfl = 2π × 50Hz × 0.5H = 157.1Ω
VL = I×XL
I = VL/XL = 110V/157.1Ω = 0.7A
(d) Define the term ‘e.m.f. of self-inductance’. Give ONE application of e.m.f. of self-
induction in electronic circuits. (5 marks)
The term "e.m.f. of self-inductance" refers to the voltage induced in a circuit due to
a change in the current flowing through a coil or inductor.
Application
It is used in switching regulators
2014 Question 7: R – L Circuit
(b) (i) State the names of the TWO resistive effects in an inductive a.c. circuit. (2 marks)
1. Resistance of the conductor
2. Eddy current losses
 (ii) Name the TWO types of electrical power in an inductive a.c. circuit. (2 marks)
1. Active power
2. Reactive power
(iii) State the name of ONE resistive component in an inductive a.c. circuit. (1 mark)
A resistor
(c) Figure 5 shows the circuit diagram of a resistor and an inductor connected in series to an
a.c. supply.

Calculate the
(i) Inductive reactance of the coil (3 marks)
Inductive reaction (XL) = 2πfL = 2π×50Hz×0.0127H = 3.99Ω

(ii) Total opposition to a.c. current flow (3 marks)

Impedance (Z) = √𝑹𝟐 + 𝑿𝑳𝟐 = √𝟑𝟐 + 𝟑. 𝟗𝟗𝟐 = √𝟗 + 𝟏𝟓. 𝟗𝟐 = √𝟐𝟒. 𝟗𝟐 =

4.99Ω

(iii) Total current flowing in the circuit (3 marks)

I=V/Z = 20V/4.99 = 4A

(iv) Active power in the circuit (3 marks)

Phase angle (Tan θ) = XL/R = 3.99/3 = 1.33

θ = Tan – 1 1.33 = 53.06o

True or active power, P = VI cosφ = 20V ×4A (cos53.06) = 80W × 0.6 = 48W

(v) Apparent power in the circuit. (3 marks)

Apparent power, S= VI = 20V × 4A = 80VA

2007 Questions 6: R – C Circuit
(a) a 80µF capacitor draws a current of 1.0A when 220V a.c. is connected across it.
Calculate:
(i) The frequency of the supply voltage (5 marks)
Vc = I×Xc
Xc = Vc/I = 220V/1A = 220Ω
Xc = 1/2πfc
𝟏 𝟏 𝟏
F = 𝟐𝛑×𝐗𝐜×𝐜 = 𝟐𝛑×𝟐𝟐𝟎Ω×(𝟖𝟎×𝟏𝟎−𝟔 ) = 𝟎.𝟏𝟏 = 9.1Hz

(ii) The value of resistance connected in series to reduce the current to 0.75A at the
same frequency. (7 marks)
Z = V/I = 220V/0.75A = 293.3Ω
R = √𝒁𝟐 − 𝑿𝒄𝟐 = √𝟐𝟗𝟑. 𝟑𝟐 − 𝟐𝟐𝟎𝟐 = √𝟖𝟔, 𝟎𝟐𝟒. 𝟖𝟗 − 𝟒𝟖𝟒𝟎𝟎 = √𝟑𝟕, 𝟔𝟐𝟒. 𝟖𝟗
= 193.97Ω
(iii) The phase angle of the RC series circuit. (3 marks)
𝑿𝒄 𝟐𝟐𝟎
Tan θ = = 𝟏𝟗𝟑.𝟗𝟕 = 1.134
𝑹

θ = tan -1 1.134 = 48.59o

(b) Make a neat sketch indicating the current and voltage relationship of the components in a
RC series circuit. (5 marks)
2013 Question 10: R – C Circuit
(b) A capacitance of 10 µf is connected in series with a 5Ω resistor across a 500V, 50Hz
supply
(i) Draw the circuit diagram and a phasor diagram showing the relationship between
the voltages and current in the circuit. (4 marks)

Circuit diagram Phasor diagram

(ii) Calculate the
a) Capacitive reactance of the capacitance (2 marks)
𝟏 𝟏 𝟏
Xc = 𝟐𝛑×𝐟×𝐜 = 𝟐𝛑×𝟓𝟎𝐇𝐳×(𝟏𝟎×𝟏𝟎−𝟔 ) = 𝟎.𝟎𝟎𝟑𝟏 = 322.58Ω

b) Impedance of the circuit (2 marks)

c) Z = √𝑹𝟐 + 𝑿𝒄𝟐 = √𝟓𝟐 + 𝟑𝟐𝟐. 𝟓𝟖𝟐 = √𝟐𝟓 + 𝟏𝟎𝟒, 𝟎𝟓𝟕. 𝟖𝟔 = √𝟏𝟎𝟒𝟎𝟖𝟐. 𝟖𝟓

= 322.62Ω

d) Current in the circuit (2 marks)

V=I×Z
I = V/Z = 500V/322.62Ω = 1.55A
e) Power factor of the circuit. (2 marks)

𝑿𝒄 𝟑𝟐𝟐.,𝟓𝟖
Tan θ = = = 64.516
𝑹 𝟓

θ = tan -1 64.516 = 89.11o

P.F = cos θ = cos 89.11 = 0.016
2004 Question 8: R-L-C Circuit
A series circuit consisting of a resistor, R = 10Ω, an inductor, L = 0.5H and a 100 – micro farad
capacitor connected to a 120V, 50Hz supply.

(a) Sketch the circuit (3 marks)

(b) Calculate the following

(i) Inductive reactance (4 marks)

XL = 2πfL = 2π×50Hz×0.5H = 157.08Ω

(ii) Capacitive reactance (4 marks)

𝟏 𝟏 𝟏
Xc = 𝟐𝛑×𝐟×𝐜 = 𝟐𝛑×𝟓𝟎𝐇𝐳×(𝟏𝟎𝟎×𝟏𝟎−𝟔 ) = 𝟎.𝟎𝟑𝟏 = 32.26Ω

(iii) Impedance (6 marks)

When XL greater than XC
Impedance, Z = √𝑹𝟐 + (𝑿𝑳 − 𝑿𝒄)𝟐 = √𝟏𝟎𝟐 + (𝟏𝟓𝟕. 𝟎𝟖 − 𝟑𝟐. 𝟐𝟔)𝟐 =

√𝟏𝟎𝟎 + 𝟏𝟓𝟓𝟖𝟎. 𝟎𝟑 = √𝟏𝟓𝟔𝟖𝟎. 𝟎𝟑 = 125.22Ω

(iv) Current in the circuit (3 marks)
V = IZ
I = V/Z = 120V/125.22Ω = 0.96A