Q.
In the circuit shown below, the current in the $1\Omega$ resistor is
AIPMTAIPMT 1988Current Electricity
Solution:
Applying Kirchhoff's loop law in loops 1 and 2 in the directions shown in figure we have
$6-3(i_1+i_2)-i_2=0$
......(i)
$9-2i_1+i_2-3i_1=0$
.......(ii)
Solving Eqs. (i) and (ii) we get,
$i_2=0.13 A$
Hence, the current in 1 Q resister is 0.13 A from Q to P
