21. What would be the correct equation representing Kirchhoff’s Current Law (KCL) at node a for the given network?
i1 – i2 + i3 – i4 = 0
i1 + i2 – i3 + i4 = 0
i1 – i2 – i3 + i4 = 0
i1 – i2 = 0
By applying KCL, at node a
i1 – i2 – i3 + i4 = 0
22. Find the value of i2, i4, and i5 if i1 = 3A, i3 = 1A and i6 = 1A by applying Kirchhoff’s current law
A. 2,-1,2
B. 4,-2,4
C. 2,1,2
D. 4,2,4
Answer: A
At junction a: i1-i3-i2 = 0. i2 = 2A.
At junction b: i4+i2-i6 = 0. i4 = -1A.
At junction c: i3-i5-i4 = 0. i5 = 2A.
23. In the circuit shown in the following figure, calculate the value of the unknown resistance R when the current in-branch OA is zero.
5 Ω
3 Ω
12 Ω
10 Ω
Answer.3. 12 Ω
Given the current through AO is zero,
It means node A and node O has the same potential,
Hence, VBA = VBO …. (1)
Also, VAC = VOC …. (2)
VAC = 4(3I) volts
VOC = IR
From equation (2),
1 × 2 I = IR
∴ R = 12 Ω
24. What is the value of current if a 50C charge flows in a conductor over a period of 5 seconds?
A. 5A
B. 10A
C. 15A
D. 20A
Answer: B
Current = Charge/Time.
Here charge = 50c
time = 5seconds,
so current = 50/5 = 10A.
25. The voltages developed across the 3 Ω and 2 Ω resistors shown in the figure are 6 V and 2 V respectively, with the polarity as marked. What is the power (in Watt) across the resistance having a voltage of 5 V?
5
7
10
14
Answer.1. 5 A
Let current through 3 Ω be I1 and through 2 Ω is I2 and current through 5 V is I ampere respectively as shown below
I1 = 6/3 = 2 A
I2 = 2/2 = 1 A
Apply KCL at N/W N2,
I + I2 = I1
I + 1 = 2
I = 1 A
P = 5 × 1 = 5 W
26. KCL deals with the conservation of?
A. Momentum
B. Mass
C. Potential Energy
D. Charge
Answer: D
KCL states that the amount of charge entering a junction is equal to the amount of charge leaving it, hence it is the conservation of charge.
27. Find the voltage V0 across 4 Ω resistor in the following circuit.
20 V
12 V
24 V
18 V
Answer.3. 24 V
Apply KCL at node ‘a’
i0 = 0.5i0 + 3
0.5i0 = 3
i0 = 6 A
∴v0 = 4i0 = 4× 6 = 24 V
28. KCL is applied at _________
A. Loop
B. Node
C. Both loop and node
D. Neither loop nor node
Answer: B
KCL states that the amount of charge leaving a node is equal to the amount of charge entering it, hence it is applied at nodes.
29. Find the value of current I using KCL
2 A
3 A
1 A
5 A
KCL at Node A
I1 + 1A = 3A
I1 = 2 A
KCL at node B
I2 + 2A = 6A
I2 = 4 A
KCL at node C
I3 = 4A + 2A
I3 = 6A
KCL at node D
I = 6A – 5A = 1 A
30. KCL can be applied for __________
A. Planar networks
B. Non-planar networks
C. Both planar and non-planar
D. Neither planar nor non-planar
Answer: C
KCL is applied for different nodes of a network whether it is planar or non-planar.
