Series circuit open-circuit measurement: A 20 V DC source supplies an 8 Ω resistor in series with a lamp. If the lamp is removed (leaving an open circuit at the lamp socket), what voltage will a voltmeter read across the lamp socket terminals?

Introduction / Context:
Measuring voltages in series circuits requires understanding where drops occur. When a component is removed, the circuit may become open, changing how voltage is distributed. This question reinforces the concept that with zero current, resistive drops vanish and the source appears across the open terminals.

Given Data / Assumptions:

• DC source voltage: 20 V.
• Series elements: 8 Ω resistor and a lamp originally in series.
• The lamp is removed, creating an open circuit at the lamp socket.
• Voltmeter has ideally infinite input resistance and is connected across the lamp socket.

Concept / Approach:
Ohm’s law dictates the voltage drop across a resistor is V_R = I * R. In an open circuit, the loop current I is zero. Therefore, the drop across the 8 Ω resistor becomes 0 V, and the entire source voltage appears across the open terminals (the lamp socket).

Step-by-Step Solution:
Lamp removed → circuit open → I = 0 A.Resistor drop: V_R = I * R = 0 * 8 = 0 V.KVL around loop: source 20 V must appear across the only remaining break → lamp socket sees 20 V.Voltmeter reading across lamp socket = 20 V.

Verification / Alternative check:
Model the lamp socket as two open terminals. With no closed path, current is zero. Any series resistive element has zero drop, leaving the full source across the open—consistent with standard open-circuit behavior.

Why Other Options Are Wrong:
0 V: would require a short across the socket, not an open.8 V or 12 V: arbitrary partial drops contradict zero-current condition (no drop across finite resistance with I = 0).

Common Pitfalls:
Forgetting that voltmeters do not provide a current path. A high-impedance meter does not “close” the circuit; thus no current flows and no series drop occurs.

Final Answer:
20 V