Q1. What is the relationship between current (I), voltage (V), and resistance (R) according to Ohm's Law?
Solution:
Ohm's Law states that current is equal to voltage divided by resistance, I = V / R.
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Q2. What is the total capacitance of two capacitors, C1 = 4μF and C2 = 6μF, connected in series?
Solution:
For capacitors in series, 1/C_eq = 1/C1 + 1/C2, so 1/C_eq = 1/4 + 1/6 = 5/12, thus C_eq = 2.4μF.
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Q3. If a capacitor has a capacitance of 5μF and is charged to a voltage of 10V, what is the charge stored in the capacitor?
Solution:
Charge (Q) is given by Q = C * V = 5μF * 10V = 0.05C.
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Q4. In a simple series circuit with a 12V battery and two resistors (4Ω and 8Ω), what is the total current flowing through the circuit?
Solution:
Total resistance R_total = R1 + R2 = 4Ω + 8Ω = 12Ω. Using Ohm's Law, I = V / R_total = 12V / 12Ω = 1A.
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Q5. What is the unit of electric potential difference?
Solution:
The unit of electric potential difference is the Volt (V).
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Q6. What is the equivalent resistance of two resistors, R1 = 6Ω and R2 = 3Ω, connected in parallel?
Solution:
In parallel, the equivalent resistance is given by 1/R_eq = 1/R1 + 1/R2, so 1/R_eq = 1/6 + 1/3 = 1/2, thus R_eq = 2Ω.
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Q7. If the voltage across a resistor is doubled while the resistance remains constant, what happens to the current?
Solution:
According to Ohm's Law, if voltage is doubled and resistance is constant, current will also double.
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Q8. In a circuit with a 12V battery and two resistors in series (4Ω and 8Ω), what is the voltage drop across the 8Ω resistor?
Solution:
Total resistance = 4Ω + 8Ω = 12Ω. Current I = V / R = 12V / 12Ω = 1A. Voltage drop across 8Ω = I * R = 1A * 8Ω = 8V.
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Q9. If a circuit has a total resistance of 10Ω and a current of 5A, what is the voltage across the circuit?
Solution:
Using Ohm's Law, V = I * R = 5A * 10Ω = 50V.
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Q10. What is the total current in a parallel circuit with two branches, where one branch has a resistance of 4Ω and the other has 6Ω, connected to a 12V source?
Solution:
First, find the equivalent resistance: 1/R_eq = 1/4 + 1/6 => R_eq = 2.4Ω. Then, I = V / R_eq = 12V / 2.4Ω = 5A.
