41.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.
42. 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.
43. 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.
44. 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.
45. Determine the current if a 20-coulomb charge passes a point in 0.25 seconds.
