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I'm given the following circuit: enter image description here

The bulbs are identical and A1, A2, and A3 are ammeters. A1 reads 0.6A

I'm told to find the readings of all the ammetres when the bulb labelled 'X' is removed such that a break occurs in that branch.

The answer I'm given is that A1 = 0.3A and A2 = 0.3A, while A3 = 0.0A. But why?

Is it not that since there is a break on the right most branch, that the circuit becomes a series circuit and so the current is no longer split? So A1 = 0.6A and a2 = 0.6A, while A3 = 0.0A?

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In the first instance, when bulb $X$ isn't removed, the voltage in the circuit is $V=0.6R_2$ where $$R_2=\frac{1}{\frac{1}{R_1}+\frac{1}{R_1}}=\frac{R_1}{2}$$ and hence $V=\frac{0.6R_1}{2}=0.3R_1$. When the other bulb is removed, you have $0.3R_1=IR_1$ and you can see how the result follows.

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You are assuming that the current is fixed on 0.6A, no matter the circuit structure. This is incorrect. The voltage of the battery is fixed and the current should be recalculated when the right branch is open.

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  • $\begingroup$ So I'm wrong in thinking that when the right most branch is broken, the components in the circuit are treated as if they were in series or that the parallel circuit becomes a series circuit since only one branch is not broken? $\endgroup$ – Kyzen Dec 29 '17 at 10:31
  • $\begingroup$ Your statement is correct. The parallel circuit does become a simple series circuit. You just need to re-calculate the current for this new circuit. $\endgroup$ – npojo Dec 29 '17 at 10:36
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V = IR.

The voltage is constant. The initial resistance is the effective resistance of 2 bulbs, each of resistance R, which is $R/2$. Hence 0.6 x $R/2$ = V. This means 0.3 x $R$ = V. Our final resistance is R. So this means our final current is 0.3. Hence A1 = A2 = 0.3, and A3 = 0 since it is an open circuit and no current flows through it. Hope this helps.

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