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You are watching: Why is there zero current when a lightbulb burns out

In the question below, shouldn"t the voltmeter review zero V bereason the circuit is damaged and also there is no present flowing? (the answer in the text is 12v)

When the lamp B burns out, current flowing via it becomes zero. So you might as well rerelocate it out of the circuit totally.

This, in general, does not intend that the existing in the circuit is additionally zero. However, assuming that the resistance of the voltmeter is exceptionally high (infinity, so to speak), you could say that the current is virtually zero.

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If the present in the circuit is zero, then the voltage drop throughout the lamp A becomes zero. Hence, you have the right to currently short circuit the lamp A. Thus, what you are currently left via is the voltmeter associated throughout the terminals of the battery. Assuming that the internal resistance of the battery is zero, the voltmeter would certainly review the EMF of the contact, i.e. 12 volts.

Hope this cleared your doubt! :)

With bulb \$B\$ damaged you currently have actually this circuit.

If the resistance of bulb \$A\$ is \$R_ m A\$ and also the resistance of the voltmeter is \$R_ m V\$ then the full resistance in the circuit is \$R_ m A + R_ m V\$.

The existing in the circuit \$I = dfrac 12R_ m A + R_ m V\$ and also so the potential difference throughout the voltmeter is \$V_ m V = I,R_ m V = dfrac 12,R_ m VR_ m A + R_ m V\$

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

If the cell has actually an internal resistance \$R_ m C\$ the analysis still works as lengthy as \$R_ m V gg R_ m A+R_ m C\$

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Graph profile of the voltage drop vs resistance throughout a resistor, and also the lutz-heilmann.info behind electric potential distinction
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