Current, Resistance and Ohms Law - Circuit Analysis Techniques

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Q1. If a circuit has a total resistance of 10Ω and a total current of 2A, what is the total voltage supplied by the battery?

Solution:

Using Ohm's Law, V = I * R = 2A * 10Ω = 20V.

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Q2. 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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Q3. 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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Q4. In a circuit with a 24V battery and a total resistance of 6Ω, what is the total current flowing through the circuit?

Solution:

Using Ohm's Law, I = V / R = 24V / 6Ω = 4A.

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Q5. A capacitor with a capacitance of 10μF is charged to a voltage of 5V. What is the charge stored in the capacitor?

Solution:

The charge stored in a capacitor is given by Q = C * V = 10μF * 5V = 50μC.

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Q6. 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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Q7. 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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Q8. 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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Q9. If a capacitor has a capacitance of 10μF and is charged to a voltage of 5V, what is the charge stored in the capacitor?

Solution:

The charge stored in a capacitor is given by Q = C * V = 10μF * 5V = 50μC.

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Q10. 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.