Two batteries, one of emf 18V and internal resistance 2 Ω Ω and the other of emf 12 V and internal resistance 1 Ω Ω , are connected as shown. The voltmeter V will record a reading of:
Text Solution
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It is clear that the two cells oppose each other hence, the effective emf in closed circuit is 18 – 12 = 6V and net resistance is 1 + 2 = 3 Ω Ω (because in the closed circuit the internal resistances of two cells are in series).
The current in circuit will be in direction of arrow shown in figure.
I = 
= 
= 2A
The potential difference across V will be same as the terminal voltage of either cell.
Since, current is drawn from the cell of 18 volt, hence,
V 1 = E 1 – ir 1
= 18 – (2 × 2) = 18 – 4 = 14 V
Similarly, current enters in the cell of 12V hence
V 2 = E 2 + ir 2
= 12 + 2 × 1
= 12 + 2 = 14 V
Hence, V = 14V
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