# Is current still split in a parallel circuit with two branches when one of the branches is broken? [closed]

I'm given the following circuit:

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?

## closed as off-topic by John Rennie, stafusa, Jon Custer, Yashas, Daniel GriscomJan 1 '18 at 20:00

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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.
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.