In a circuit with a 60 V source and two resistors 10 Ω and 20 Ω in series, calculate total current and voltage drops across each resistor.

In a circuit with a 60 V source and two resistors 10 Ω and 20 Ω in series, calculate total current and voltage drops across each resistor.

• A. I = 0.75 A; V_10Ω = 7.5 V; V_20Ω = 15 V
• B. I = 1 A; V_10Ω = 10 V; V_20Ω = 50 V
• C. I = 2 A; V_10Ω = 40 V; V_20Ω = 20 V
• D. I = 2 A; V_10Ω = 20 V; V_20Ω = 40 V

Answer: (D)

In a series circuit, the same current flows through every component, and the total resistance is the sum of the individual resistances. So the total resistance here is 10 Ω + 20 Ω = 30 Ω. The current drawn from the 60 V source is I = V / R = 60 / 30 = 2 A. With 2 A flowing, the voltage drop across the 10 Ω resistor is V = I × R = 2 × 10 = 20 V, and across the 20 Ω resistor is V = I × R = 2 × 20 = 40 V. These drops add up to the 60 V supply, as expected. This matches the correct result: current 2 A, 10 Ω drop 20 V, 20 Ω drop 40 V. The other options conflict with either the current or the individual drops given the series arrangement and Ohm’s law.

In a series circuit, the same current flows through every component, and the total resistance is the sum of the individual resistances. So the total resistance here is 10 Ω + 20 Ω = 30 Ω. The current drawn from the 60 V source is I = V / R = 60 / 30 = 2 A.

With 2 A flowing, the voltage drop across the 10 Ω resistor is V = I × R = 2 × 10 = 20 V, and across the 20 Ω resistor is V = I × R = 2 × 20 = 40 V. These drops add up to the 60 V supply, as expected.

This matches the correct result: current 2 A, 10 Ω drop 20 V, 20 Ω drop 40 V. The other options conflict with either the current or the individual drops given the series arrangement and Ohm’s law.