Q1. What is the equivalent impedance (Z) of a 4Ω resistor in series with a 3Ω resistor?
Solution:
Z = R1 + R2 = 4Ω + 3Ω = 7Ω.
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Q2. If a circuit has a total impedance of 4Ω and a current of 3A, what is the voltage across the circuit?
Solution:
Using Ohm's law, V = I * Z = 3A * 4Ω = 12V.
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Q3. What is the Thevenin equivalent voltage of a circuit with a 10V source and a 5Ω resistor in series with a 10Ω resistor?
Solution:
Thevenin equivalent voltage is the open-circuit voltage, which is 10V across the 5Ω resistor.
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Q4. If a circuit has a Thevenin equivalent of 15V and a load resistance of 5Ω, what is the load current?
Solution:
Using Ohm's law: I = V/R = 15V / 5Ω = 3A.
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Q5. What is the Thevenin equivalent voltage across a load resistor in a simple circuit with a 12 V source and a 4 ohm resistor in series with a 2 ohm load?
Solution:
The Thevenin equivalent voltage is the voltage across the load resistor, which can be found using voltage division: V_load = V_source * (R_load / (R_series + R_load)) = 12 V * (2 / (4 + 2)) = 8 V.
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Q6. If a circuit has a total voltage of 120V and a total current of 10A, what is the total power in the circuit?
Solution:
Power P = V * I = 120V * 10A = 1200W.
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Q7. What is the Thevenin equivalent voltage of a circuit with a 10 V source and a 5 ohm resistor in series with a 10 ohm load?
Solution:
The Thevenin equivalent voltage is the open-circuit voltage, which is 10 V in this case.
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Q8. What is the Thevenin equivalent voltage across terminals A and B if a 12V battery is connected in series with a 4Ω resistor?
Solution:
The Thevenin equivalent voltage is the open-circuit voltage across terminals A and B, which is equal to the source voltage, 12V.
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Q9. In a parallel circuit with a 12V source and two resistors (4Ω and 12Ω), what is the total current supplied by the source?
Solution:
Total current I = V * (1/R1 + 1/R2) = 12V * (1/4 + 1/12) = 3A.
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Q10. In a series circuit with a 9V battery and two resistors (3Ω and 6Ω), what is the voltage across the 6Ω resistor?
Solution:
Using voltage division, V = V_total * (R / (R1 + R2)) = 9V * (6Ω / (3Ω + 6Ω)) = 6V.
