Ammeter and voltmeter question?

In the circuit shown in Figure, all meters are idealized and the batteries have no appreciable internal resistance.

Answer 1

When the switch is open:
Vab+15-(25V+15V)/(100Ω+75Ω)*75 = 0 so Vab = -15V + (25V+15V)/(100Ω+75Ω)*75Ω = 2.14V

The ideal ammeter is a short circuit so when the switch is closed you just draw a short from a to b. Obviously Vab = 0 and voltmeter reads zero when the switch is closed.
Then by KVL you know the 100Ω has 25V across it meaning it has 25/100 = 0.25A left to right. Likewise the 75Ω by KVL must have 15V across it and 15V/75Ω = 0.2A down to the bottom of the ckt. So you have 250ma entering the ammeter from above and 200ma leaving that same node right to the 15V source. That leaves 250ma – 200ma = 50ma into the ammeter. We know the ammeter shorts out the voltmeter so the voltmeter reads 0V and the ammeter reads 50ma = 0.05A = 1/20 A
Draw the ckt with and without the switch closed and then write in the currents in the branches and verify the KCL (current entering = current leaving) and KVL (sum of voltage drops and rises around a closed loop = 0) for further understanding. Hope this helps!

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