Solution:
The potential difference is the equivalent EMF: \(E_{\text{eq}} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{18/3 + 12/2}{1/3 + 1/2} = \frac{12}{5/6} = 14.4\text{ V}\).
Two batteries one of emf \(18\text{ V}\) and internal resistance \(3\text{ }\Omega\) while other of emf \(12\text{ V}\) and internal resistance \(2\text{ }\Omega\) are connected in parallel with positive terminals together at one point and negative terminals together to other point. If across these points an ideal voltmeter is connected, the reading of voltmeter will be
Leave a Reply