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If Lamp2 is disconnected, What will happen to the brightness of the Lamp3?

I think that the brightness will remain the same because Lamp2 is in parallel with the branch that contains Lamp3.

Let's assume that I3 passes in Lamp3. When Lamp2 is disconnected, The total current will be reduced to be equal to I3. So Lamp3 and Lamp4 will not be affected. That's why loads are connected in parallel at home.

This question is in my younger brother's scholar book. The answer was: "The brightness of Lamp3 will increase because the total current will pass through it."

I think I'm missing something but I don't know what it is.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ is that battery current limited? \$\endgroup\$ – JonRB May 5 '17 at 13:48
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    \$\begingroup\$ You haven't considered lamp 1 at all. Consider lamp 1. \$\endgroup\$ – Dampmaskin May 5 '17 at 13:49
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    \$\begingroup\$ Doesn't matter. Suppose battery has voltage U and lamp has resistance R \$\endgroup\$ – Gregory Kornblum May 5 '17 at 14:02
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    \$\begingroup\$ You don't need numbers -- you simply assume that that all four lamps are identical in problems like this. \$\endgroup\$ – Dave Tweed May 5 '17 at 14:03
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    \$\begingroup\$ You've got a reputation of 1,686 and you can't answer THIS?? \$\endgroup\$ – Finbarr May 5 '17 at 16:01
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You are ignoring lamp 1... In this instance you can consider Lamp 1 as the internal resistance of the battery.

That is, the current through Lamp2 PLUS the series chain Lamp3/4 MUST be drawn through Lamp1 with an attendant voltage drop across Lamp1.

When you remove Lamp2, less current is drawn through Lamp1. Less voltage is dropped across it, and so more current is drawn by Lamp3/4. Till it rebalances.

In the end Lamp1 gets dimmer and Lamp3/4 get brighter.

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When you turned​ off bulb2, total current flowing from the battery get decrease and while blub1 in series it will dim and voltage drop upon it also decrease. As the kirchoff's law total voltage drop each resistor in series are same this decreased voltage drop on bulb1 increse the voltage drop on bulb2 and bulb3 increase so these two get brighten.

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