Network theory miscellaneous
Direction: 20 to Q. 25. For the circuit shown below—
- Power delivered by 6A current source—
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From figure given in solution 21:
Voltage across terminal D and C is 1V
since terminal D is negative, therefore power delivered by 6A current source is negative
Pdelivered = – VI = – 1 × 6 = – 6W.Correct Option: B
From figure given in solution 21:
Voltage across terminal D and C is 1V
since terminal D is negative, therefore power delivered by 6A current source is negative
Pdelivered = – VI = – 1 × 6 = – 6W.
- The power delivered by 8A current source—
-
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From above figure:
Voltage across terminals A and C is 30 – 4 = 26V; since terminal A is positive, therefore power delivered by the 8A current source is positive
Pdelivered = VI = 26 × 8 = 208W.Correct Option: A
From above figure:
Voltage across terminals A and C is 30 – 4 = 26V; since terminal A is positive, therefore power delivered by the 8A current source is positive
Pdelivered = VI = 26 × 8 = 208W.
Direction: Statement for Q. 1 to Q. 3.
For the circuit shown below, it is known that I2 = 3A.
- RL = ?
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RL = Vin = 24V = 12Ω I3 2A Correct Option: A
RL = Vin = 24V = 12Ω I3 2A
- Vin = ?
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Vin = I2 × 8Ω = 3 × 8 = 24V.
Correct Option: C
Vin = I2 × 8Ω = 3 × 8 = 24V.
- I3 = ?
-
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I1 = Vin = 24V = 8A 3Ω 3Ω
∴ 12 = I1 + I2 + I3
so
I3 = 12 – I1 – I2 = 12 – 8 – 2 = 2ACorrect Option: A
I1 = Vin = 24V = 8A 3Ω 3Ω
∴ 12 = I1 + I2 + I3
so
I3 = 12 – I1 – I2 = 12 – 8 – 2 = 2A
