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How does a parallel circuit change when a branch is added?
The total resistance increases, so the current in the circuit decreases.
The total resistance decreases, so the current in the circuit increases.
The total resistance increases, so bulbs shine at a lower brightness.
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Correct answer: "The total resistance decreases, so the current in the circuit increases."
Why:
- For resistors in parallel the conductances add: 1/R_total = 1/R1 + 1/R2 + ... + 1/Rn, so adding another branch lowers R_total.
- With a fixed supply voltage V, I_total = V / R_total therefore I_total increases when R_total falls.
- The current through each existing branch (and thus the brightness of bulbs in those branches) stays essentially the same because the voltage across each parallel branch remains the supply voltage.
Caveat:
- If the source has significant internal resistance (or the supply voltage droops under higher load), adding a branch can reduce the terminal voltage and so could dim existing bulbs. With an ideal fixed-voltage source, bulb brightness does not change.
Why:
- For resistors in parallel the conductances add: 1/R_total = 1/R1 + 1/R2 + ... + 1/Rn, so adding another branch lowers R_total.
- With a fixed supply voltage V, I_total = V / R_total therefore I_total increases when R_total falls.
- The current through each existing branch (and thus the brightness of bulbs in those branches) stays essentially the same because the voltage across each parallel branch remains the supply voltage.
Caveat:
- If the source has significant internal resistance (or the supply voltage droops under higher load), adding a branch can reduce the terminal voltage and so could dim existing bulbs. With an ideal fixed-voltage source, bulb brightness does not change.
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