question_answer

A 7 V battery with internal resistance \[2\,\Omega \] and a 3 V battery with internal resistance \[1\,\Omega \] are connected to a \[10\,\Omega \] resistor as shown in figure, the current in \[10\,\Omega \] resistor is:

A) 0.27 A                

B) 0.31 A

C) 0.031 A               

D) 0.53 A

Correct Answer: C

Solution :

(c) 0.031 A Using Kirchhoff's law in loop \[A\,{{P}_{2}}{{P}_{1}}DA\] \[\therefore 10{{l}_{1}}+2l-7=0\]             \[10{{l}_{1}}+2l=7\]                …(1) Using Kirchhoff's law in loop \[{{P}_{2}}{{P}_{1}}CB{{P}_{2}}\]             \[-3+1(l-{{l}_{1}})-10{{l}_{1}}=0\]             \[l-11{{l}_{1}}=3;l=3+11{{l}_{1}}\]    ….(2) From eqs. (1) and (2), we get \[10{{l}_{1}}+2(3+{{11}_{1}})=7\]             \[\Rightarrow 10{{l}_{1}}+6+22{{l}_{1}}=7\]             \[\therefore 32{{l}_{1}}=1:{{l}_{1}}=\frac{1}{32}=0.031A\]