Series drop with a lamp: An 8 Ω resistor is in series with a lamp. The circuit current is 1 A and the applied supply is 20 V DC. What voltage is available across the lamp?

Introduction / Context:
Series circuits share the same current, but sources divide their voltage among components according to their resistances. This practical problem shows how a series “dropping resistor” reduces the voltage applied to a lamp—useful for quick design checks and troubleshooting.

Given Data / Assumptions:

• Resistor R = 8 Ω in series with a lamp.
• Current I = 1 A through the series path.
• Supply voltage V_s = 20 V DC.
• Lamp modeled as a load that takes the remaining voltage.

Concept / Approach:
Use Ohm’s law to find the resistor’s voltage drop, then subtract from the source to find the lamp voltage. In any series loop, V_s = V_R + V_lamp. With known I and R, V_R = I * R.

Step-by-Step Solution:

Verification / Alternative check:
Power check: P_R = I^2 * R = 1^2 * 8 = 8 W; P_lamp = V_lamp * I = 12 * 1 = 12 W; total 8 W + 12 W = 20 W = V_s * I, confirming consistency.

Why Other Options Are Wrong:

• 4 V or 8 V: 8 V is the resistor drop, not the lamp. 4 V is unsupported by the data.
• 20 V: Would be true only with no series resistor drop (not our case).

Common Pitfalls:
Forgetting that the same current flows through both elements; mixing up which voltage belongs to which component.

Final Answer:
12 V