question_answer

Two sources of emf 6V and internal resistance \[3\,\Omega \] and \[2\Omega \] are connected to an external resistance R as shown. If potential difference across battery A is zero then value of R is -

A) \[1\Omega \]                 

B) \[2\Omega \] 

C) \[3\Omega \]                             

D) \[4\Omega \]

Correct Answer: A

Solution :

[A]

Current in the circuit is i and \[i=\frac{{{\in }_{eq}}}{R+{{r}_{eq}}}\]

\[{{\in }_{eq}}=6+6=12,\] \[{{r}_{eq}}=3+2=5\]

\[i=\frac{12}{R+5}\]

p.d. across A=0

\[{{\in }_{A}}-i{{r}_{A}}=0,\] where \[{{\in }_{A}}=6V\] and \[{{r}_{A}}=3\Omega \]

\[6-\frac{12}{R+5}\times 3=0\]\[R=1\Omega \]