JEE Advanced · Physics · 19. Current Electricity
Two batteries of different emfs and different internal resistances are connected as shown. The voltage across \(A B\) in volt is

- A 5
- B 10
- C 15
- D 12
Answer & Solution
Correct Answer
(A) 5
Step-by-step Solution
Detailed explanation
\(V_{A B}=\) Equivalent emf of two batteries in parallel.
\(
\begin{aligned}
& =\frac{E_1 / r_1+E_1 / r_2}{1 / r_1+1 / r_2} \\
& =\frac{(6 / 1)+(3 / 2)}{(1 / 1)+(1 / 2)}=5 \mathrm{~V}
\end{aligned}
\)
\(\therefore\) Answer is 5 .
Analysis of Question
(i) In my opinion this is the simplest question of JEE 2011.
(ii) When no current is drawn from this equivalent battery, then
\(
V_{A B}=V=E
\)
Otherwise, \(V=E \pm\) ir
(iii) Similar type of question was asked in JEE 1981 and cancelled paper of JEE 1997.
\(
\begin{aligned}
& =\frac{E_1 / r_1+E_1 / r_2}{1 / r_1+1 / r_2} \\
& =\frac{(6 / 1)+(3 / 2)}{(1 / 1)+(1 / 2)}=5 \mathrm{~V}
\end{aligned}
\)
\(\therefore\) Answer is 5 .
Analysis of Question
(i) In my opinion this is the simplest question of JEE 2011.
(ii) When no current is drawn from this equivalent battery, then
\(
V_{A B}=V=E
\)
Otherwise, \(V=E \pm\) ir
(iii) Similar type of question was asked in JEE 1981 and cancelled paper of JEE 1997.
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