What is the voltage drop across R1?

Source voltage = 12 Volts

R1 = 5 Ohm

R2 = 15 Ohms

Question

What is the voltage drop across R1?

Source voltage = 12 Volts

R1 = 5 Ohm

R2 = 15 Ohms

BlackTom AI · Accepted Answer

The circuit shows two resistors in series connected to a 12 V source. In a series circuit, the same current flows through both resistors, and the voltage divides proportionally to resistance.
Option 6 Volts: If R1 dropped 6 V, the current would have to satisfy V = I*R1 → I = 6 V / 5 Ω = 1.2 A. In a series circuit with R1 and R2, the total resistance would be 5 Ω + 15 Ω = 20 Ω, which would then require I = 12 V / 20 Ω = 0.6 A. This inconsistency shows 6 V cannot be the drop across R1.
Option 3 Volts: Using the series current I = 12 V / (R1 + R2) = 12 V / (5 Ω + 15 Ω) = 12 V / 20 Ω = 0.6 A, the voltage across R1 is V_R1 = I * R1 = 0.6 A * 5 Ω = 3 V. This matches the expected drop for R1 in this configuration.
Option 9 Volts: If R1 dropped 9 V, the current would be I = 9 V / 5 Ω = 1.8 A. The total circuit current for the given 12 V source would then imply a total resistance of 12 V / 1.8 A ≈ 6.67 Ω, which contradicts the actual series resistance of 20 Ω. Hence 9 V is not feasible here.
Thus, the only consistent voltage drop across R1 in this series circuit is 3 Volts.

What is the voltage drop across R1? Source voltage = 12 Volts R1 = 5 Ohm R2 = 15 OhmsSingle choice

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