(a) In the first case, the voltmeter is of high-resistance so we can say that it is an ideal voltmeter.
Potential across the terminal measured by voltmeter, E = 6.3 V
So, the emf of the battery = 6.3 V
(b) Now the current measured by the ammeter, I = 150 mA = 0.150 A
New voltage = 5.9 V
So, the voltage drop, = 6.3 - 5.9 = 0.40 V
This voltage drop is due to the internal resistance of the battery.
Therefore, the internal resistance of the battery, r = / I = 0.40 / 0.150 = 2.67
Hope you understand the solution !
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