The Attempt at a Solution
Voltmeter can be replaced by an open circuit and ammeter can be replaced by a short circuit .By doing so the current flows in the outer loop consisting of both the batteries and both the resisters .The current in the circuit is given by i = (45-15)/(20+10) = 1A .
Potential drop across Voltmeter is U2-iR2 = 45-(1)(20) = 25 V
So reading of ammeter is 1A and that of voltmeter is 25V .
Is that the correct answer for part a) ?
I am having some trouble with the use of "ideal voltmeter" .An ideal voltmeter has infinite resistance .It can be replaced by an open circuit . But if voltmeter has to read a potential difference it must draw some current .If no current flows through the voltmeter ,then it's reading would be simply zero .
Sorry for being a bit confused .