• 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) $4\Omega$

[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$