# Voltage when lamp burns out [closed]

In the question below, shouldn't the voltmeter read zero V because the circuit is broken and there is no current flowing? (the answer in the text is 12v) ## closed as off-topic by Aaron Stevens, John Rennie, Gert, ZeroTheHero, Jon CusterFeb 17 at 5:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, John Rennie, Gert, ZeroTheHero, Jon Custer
If this question can be reworded to fit the rules in the help center, please edit the question.

When the lamp B burns out, current flowing through it becomes zero. So you might as well remove it out of the circuit completely.

This, in general, does not mean that the current in the circuit is also zero. However, assuming that the resistance of the voltmeter is extremely high (infinity, so to speak), you could say that the current is nearly zero.

If the current in the circuit is zero, then the voltage drop across the lamp A becomes zero. Hence, you can now short circuit the lamp A. Thus, what you are now left with is the voltmeter connected across the terminals of the battery. Assuming that the internal resistance of the battery is zero, the voltmeter would read the EMF of the call, i.e. 12 volts.

Hope this cleared your doubt! :)

With bulb $$B$$ broken you now have this circuit.

If the resistance of bulb $$A$$ is $$R_{\rm A}$$ and the resistance of the voltmeter is $$R_{\rm V}$$ then the total resistance in the circuit is $$R_{\rm A} + R_{\rm V}$$.

The current in the circuit $$I = \dfrac {12}{R_{\rm A} + R_{\rm V}}$$ and so the potential difference across the voltmeter is $$V_{\rm V} = I\,R_{\rm V} = \dfrac {12\,R_{\rm V}}{R_{\rm A} + R_{\rm V}}$$

If the resistance of the voltmeter is much larger than the resistance of bulb $$A$$, $$R_{\rm V} \gg R_{\rm A}$$, then $$V_{\rm V} \approx 12$$ the emf of the cell.

If the cell has an internal resistance $$R_{\rm C}$$ the analysis still works as long as $$R_{\rm V} \gg R_{\rm A}+R_{\rm C}$$