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I was wondering if there is a break in one branch, will the potential difference across it still be equal to that of the branches parallel to it? I know that the top wire would still have to be at the same potential as the other top wires because the wire is an equipotential, and the same idea applies to the bottom wire, which would give it the same potential difference across the branch. However, if no current can flow across the resistor, where would the potential difference come from? Also, would anything different happen if instead it were a capacitor in the broken branch instead of a resistor (like in an RC circuit when the capacitor is almost completely charged and acts like a break in the circuit)? Thank you!

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The potential across each resistance is the same, that which is established by the battery (it's EMF). This ignores any resistance in the wires. The break in the middle resistor wire in your diagram simply causes that path to act like a resistor with infinite resistance; again, the same as measuring the potential directly across the battery.

The voltage across a single capacitor replacing one of these resistors would also be equal to the voltage across the battery. Again, the EMF of the battery and ignoring any resistance in the wires.

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