Q1. If a circuit has a Norton equivalent current of 3A and a Norton equivalent resistance of 4Ω, what is the equivalent voltage?
Solution:
Using Ohm's Law, V = I * R = 3A * 4Ω = 12V.
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Q2. What is the equivalent impedance (Z) of a circuit with a 3Ω resistor and a 4Ω inductor in series at a frequency where the inductive reactance is 4Ω?
Solution:
Z = R + jX = 3Ω + j4Ω; |Z| = √(3^2 + 4^2) = 5Ω.
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Q3. What is the total power dissipated in a circuit with a 24V source and a total resistance of 8Ω?
Solution:
Power P = V^2 / R = 24V^2 / 8Ω = 576 / 8 = 72W.
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Q4. In an AC circuit, what is the phase difference between voltage and current in a purely resistive circuit?
Solution:
In a purely resistive circuit, the voltage and current are in phase, resulting in a phase difference of 0 degrees.
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Q5. Using KCL, if three currents entering a node are 5A, 3A, and 2A, what is the total current leaving the node?
Solution:
According to KCL, total current entering = total current leaving. 5A + 3A + 2A = 10A leaving.
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Q6. What is the total resistance in a parallel circuit with two resistors, R1 and R2?
Solution:
The total resistance in a parallel circuit is given by the formula 1 / (1/R1 + 1/R2).
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Q7. If the total power in a circuit is 100W and the power factor is 0.8, what is the apparent power?
Solution:
Apparent power (S) is calculated using S = P / power factor. Here, S = 100W / 0.8 = 125VA.
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Q8. In a series circuit, how does the total current relate to the individual branch currents according to KCL?
Solution:
According to Kirchhoff's Current Law (KCL), the total current entering a junction is equal to the total current leaving the junction, which means the total current is the sum of the branch currents.
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Q9. In an AC circuit, what does the impedance (Z) represent?
Solution:
Impedance (Z) is the total opposition that a circuit offers to the flow of alternating current, combining both resistance and reactance.
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Q10. In a parallel circuit with two resistors, R1 = 4Ω and R2 = 6Ω, what is the total resistance?
Solution:
Total resistance R = 1 / (1/R1 + 1/R2) = 1 / (1/4 + 1/6) = 2.4Ω.
