Network Elements and the Concept of Circuit
- The each branch of circuit graph shown below represent a circuit element. The value of voltage V1 is—
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Apply KVL in loop ABCJKLA
100 – 105 + 15 – 55 + V1 + 30 = 0
V1 = 105 + 55 – 100 – 15 – 30
or V1 = 160 – 145 = 15 VCorrect Option: D
Apply KVL in loop ABCJKLA
100 – 105 + 15 – 55 + V1 + 30 = 0
V1 = 105 + 55 – 100 – 15 – 30
or V1 = 160 – 145 = 15 V
- Find V0 for the circuit given below—
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Voltage is constant because of 15 V source.
Correct Option: B
Voltage is constant because of 15 V source.
- Calculate the value of Vx for the circuit given below—
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To calculate VX, apply KVL in loop ABCDA we get
– VX– 8 + 6 + 5 = 0
or VX = 11 – 8 = 3 VCorrect Option: A
To calculate VX, apply KVL in loop ABCDA we get
– VX– 8 + 6 + 5 = 0
or VX = 11 – 8 = 3 V
- The voltage V0 shown below is always equal to—
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It is not possible to determine the voltage across 4 A source.
Correct Option: D
It is not possible to determine the voltage across 4 A source.
Direction: Figure given below shows four light bulbs connected across an 85V battery. The bulbs are said to be connected in parallel.
- The total power consumed in the circuit shown in the figure is—
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The given circuit
Case 1: When current source is taken
IAB = 2A & IBC = OA (Due to S.C. of voltage source)
Case 2: When voltage source is takenIBC = 2 = 1 A 2
IBA = OA
So, the power consumed in the circuit
P = I2AB × 2Ω + = I2BC × 2Ω
= 22 × 2 + 12 × 2 = 10WCorrect Option: B
The given circuit
Case 1: When current source is taken
IAB = 2A & IBC = OA (Due to S.C. of voltage source)
Case 2: When voltage source is takenIBC = 2 = 1 A 2
IBA = OA
So, the power consumed in the circuit
P = I2AB × 2Ω + = I2BC × 2Ω
= 22 × 2 + 12 × 2 = 10W
