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For the parallel circuit below:

enter image description here

Why is the current across the ammeter unchanged when the resistance of the variable resistor is increased?

I've always learnt that current varies in parallel and voltage is constant across each 'branch'. So when the variable resistor's resistance is increased, wouldn't the:

  1. (this I know right) Voltmeter reading decrease since it has less proportion of resistance in its branch.

  2. Ammeter reading increase since current would rather flow to that branch which has less resistance?

However the answer is actually that the ammeter reading is unchanged.

Thanks.

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1 Answer 1

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As you said,

voltage is constant across each 'branch'

then, remember that the current $I$ is:

$I=V/R$

so the current did not change in that branch. The total current will decrease though, because the current through the other branch did diminish (same equation, but $R $ increases).

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  • $\begingroup$ Oh so instead of the lost current through branch 1 (varied R) being diverted to branch 2, instead the total current is reduced? I didn't know that the total current could decrease...this is happening because the total R in the circuit decreases right? So if the increased resistance was compensated with loss of R in the branch with the ammeter, would the total current be the same and the ammeter reading increase? $\endgroup$
    – user9856
    Commented Nov 22, 2014 at 8:57
  • $\begingroup$ yes, but just notice the decreased R required will not be equal to the increased R because they are in parallel, so you need to calculate how much R should decrease to keep the total current constant. $\endgroup$
    – user65081
    Commented Nov 22, 2014 at 9:02
  • $\begingroup$ "this is happening because the total R in the circuit decreases right?" I guess you meant increase instead of decrease, right? $\endgroup$
    – user65081
    Commented Nov 22, 2014 at 9:04
  • $\begingroup$ Yup sorry about that. And the R thing would be because the formula is 1/R + 1/R..= 1/Rtotal right? (instead of adding them up like in series) $\endgroup$
    – user9856
    Commented Nov 22, 2014 at 9:05
  • $\begingroup$ yes. .......... . $\endgroup$
    – user65081
    Commented Nov 22, 2014 at 9:06

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