Two batteries, one of emf 18 V and internal resistance \[2\,\,\Omega \] and the other of emf 12 V and internal resistance \[1\,\,\Omega ,\] are connected as shown. The voltmeter V will record a reading of: [AIPMT (S) 2005] |
A) 15 V
B) 30 V
C) 14 V
D) 18 V
Correct Answer: C
Solution :
[c] It is clear that the two cells oppose each other hence, the effective emf in closed circuit is \[182=6\text{ }V\] and net resistance is \[1+2=3\Omega \] (because in the closed circuit the inters. resistances of two cells are in series). |
The current in circuit will be in direction of arrow shown in figure. |
\[I=\frac{\text{effective}\,\text{emf}}{\text{total}\,\text{resistance}}=\frac{6}{3}=2\,A\] |
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, |
\[{{V}_{1}}={{E}_{1}}-i{{r}_{1}}\] |
\[=18-(2\times 2)=18-4=14\,V\] |
Similarly, current enters in the cell of 12 V, hence, |
\[{{V}_{2}}={{E}_{2}}+i{{r}_{2}}\] |
\[=12+2\times 1\] |
\[=12+2=14\,V\] |
Hence, \[V=14\text{ }V\] |
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