21. Kirchhoff’s Current law is based on the law of conservation of _________
A. energy
B. momentum
C. mass
D. charge
Answer: D
Kirchhoff’s current law is based on the law of conservation of charge i.e, charge that flows in = charge that flows out.
22. The figure shown here is a branch of an electric current where the current is moving from A to B. Find the value of (VA – VB).
5 V
30 V
20 V
10 V
Answer.4. 10 V
By applying the KVL,
-VA + 2(5) + 10 – 16 + 2(3) + VB = 0
⇒ VA – VB = 10 V
23. The current law represents a mathematical statement of fact that ________
A. voltage cannot accumulate at node
B. charge cannot accumulate at node
C. charge at the node is infinite
D. none of the mentioned
Answer: B
Kirchhoff’s current law is based on the law of conservation of charge i.e, charge that flows in = charge that flows out. The charge cannot accumulate at the node, it can only flow in and out of the node.
24. In the circuit shown below, the voltage and current sources are ideal. The voltage Vout across the current source, in volts, is
0
5
10
20
Answer.4. 20
Apply KVL in the loop as shown below,
The current flow through 2 Ω is 5 A as shown in fig and it will give a drop of 10 V.
– Vout + 10 + 10 = 0
Vout = 20 V
25. Kirchhoff’s current law is applied at ___________
A. loops
B. nodes
C. both loop and node
D. none of the mentioned
Answer: B
Kirchhoff’s current law is based on the law of conservation of charge i.e, charge that flows in = charge that flows out.
Kirchhoff’s current law can be applicable to nodes only.
26. In the circuit shown below, what will be the power absorbed by the dependent source (voltage source)?
Power absorbed by dependent voltage source = V × I
= 5 v0 × I
= 5 × 0.5 × 0.1
= 0.25 W
27. Determine the current in all resistors in the circuit shown below.
A. 2A, 4A, 11A
B. 5A, 4.8A, 9.6A
C. 9.3A, 20.22A, 11A
D. 10.56A, 24.65A, 14.79A
Answer: D
All the resistors are in parallel, so the voltage across each resistor is the same V.
i1 = V/7
i2 = V/3
i3 = V/5.
By current law
50A = V/7 + V/3 + V/5.
On solving, we obtain V and then values of i1, i2, and i3.
28.The diode shown in the Figure below has zero cut-in voltage and zero forward resistance. The diode current io is
– 4 A
0 A
1 A
4 A
Answer.3. 1 A
Consider V is the node voltage,
Applying KCL to the node,
I = I1 + io …. (1)
io = I2 – 2 …. (2)
From equations (1) and (2),
I = I1 + I2 – 2 …. (3)
I = 10 – V/4
I1 = V1/4
I2 = V/1 = V
Put the value currents in equation (3),
\(\frac{{10 – V}}{4} = \frac{V}{4} + V – 2\)
3V/2 = 4.5
V = 3 volts
I2 = V = 3 A
From equation (2),
io = I2 – 2 = 3 – 2
io = 1 A
Hence diode current will be 1 A.
29. For the circuit below, find the voltage across the 5Ω resistor and the current through it.
A. 1.93 V
B. 2.83 V
C. 3.5 V
D. 5.7 V
Answer: B
Here all the resistors are connected in parallel and let the voltage be V. Hence,
i15 = V/15,
i5 = V/5,
i2 = V/2,
i1 = V/1.
By Kirchhoff’s current law,
V/15 + V/5 + V/2 V/1 +5 = 10.
V = 2.83 V
By solving an equation, we obtain the value of V.
As all resistors are in parallel, the voltage across each is the same as V.
30. Determine the current through the resistor R3 shown in the figure using KCL.
A. 25mA
B. 10mA
C. 20mA
D. 35mA
Answer: A
According to Kirchhoff’s current law (KCL), the algebraic sum of the electric currents meeting at a common point is zero. I.e. the sum of currents entering a node is equal to the sum of currents leaving the node. It is based on the conservation of charge.
Using KCL,
60mA = 10mA + 25mA + i3.
i3 = 25mA
31. Kirchhof’s current law can be mathematically stated as ___________
A. ∑k = 0n I = 0
B. i2∑k = 0n I = 0
C. i∑k = 0n I = 0
D. none of the mentioned
Answer: A
KCL states that the sum of currents entering and leaving a node is equal to zero.
32. Determine the current if a 20-coulomb charge passes a point in 0.25 seconds.
