Question 2
(a) State:
(i) Kirchhoff’s current law;
(ii) Kirchhoff’s voltage law.
Fig. 1 is a d.c. electric circuit.
(b) Calculate the currents:
(i) I1;
(ii) I2;
(iii) I3.
The expected answers were:
(a)(i) Kirchhoff’s Current Law states that the sum of the currents entering a junction is equal to the sum of the currents leaving the junction.
(ii) Kirchhoff’s Voltage Law states that the voltage applied to a closed circuit
equals the sum of the voltage drops in that circuit.
(b)(i) Using KVL,
Loop ABEFA,
10 I1 + 5 I3 = 5 ………………………………………(1)
Loop BCDEB,
15 I2 – 5 I3 = 2 ……………………………………….(2)
Using KCL,
I1 = I2 + I3
I3 = I1 – I2 ………………………………………..(3)
Substituting (3) into (1) and (2),
10 I1 + 5 (I1 – I2) = 5
15 I1 – 5 I2 = 5
3 I1 - I2 = 1 ………………………………………..(4)
15 I2 – 5 (I2 – I1) = 2
20 I2 - 5 I1 = 2 ………………………………………..(5)
Substitute from (4) I2 = 3 I1 – 1 into (5)
20 (3 I1 – 1) – 5 I1 = 2
60 I1 – 20 – 5 I1 = 2
55 I1 = 22
I1 = = 0.4 A
Substitute for I1 in (4)
I2 = 3 x 0.4 – 1 = 0.2 A
and I3 = I1 – I2 = 0.4 – 0.2 = 0.2 A
I3 = I1 – I2 = (0.32 -0.36) A = - 0.04 A
The question was on Basic Electricity. The question was unpopular with the candidates. Many candidates could not state Kirchhoff’s laws correctly which also spelt their undoing in part (b). The candidates’ performance was not encouraging. |