In a circuit with a 12 V supply and two resistors in series, 3 Ω and 9 Ω, what is the current and voltage drop across each resistor?

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Multiple Choice

In a circuit with a 12 V supply and two resistors in series, 3 Ω and 9 Ω, what is the current and voltage drop across each resistor?

In a series circuit, the same current flows through every component, and the total resistance is the sum of the individual resistances. Here, the resistors are 3 Ω and 9 Ω, so the total resistance is 12 Ω. With a 12 V supply, the current is 12 V ÷ 12 Ω = 1 A. The voltage drop across each resistor is found from V = I × R. So the 3 Ω resistor drops 1 A × 3 Ω = 3 V, and the 9 Ω resistor drops 1 A × 9 Ω = 9 V. The drops add up to the supply voltage (3 V + 9 V = 12 V), which confirms the calculations.

In a series circuit, the same current flows through every component, and the total resistance is the sum of the individual resistances. Here, the resistors are 3 Ω and 9 Ω, so the total resistance is 12 Ω. With a 12 V supply, the current is 12 V ÷ 12 Ω = 1 A. The voltage drop across each resistor is found from V = I × R. So the 3 Ω resistor drops 1 A × 3 Ω = 3 V, and the 9 Ω resistor drops 1 A × 9 Ω = 9 V. The drops add up to the supply voltage (3 V + 9 V = 12 V), which confirms the calculations.

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