Discussion Forum



1)

In the circuit  shown below, the current in the 1Ω resistor is 

2622021541_m24.PNG


A) 1.3 A, from P to Q

B) 0 A

C) 0.13 A , from Q to P

D) 0.13 A , from P to Q

Answer:

Option C

Explanation:

Central Idea: Connect point Q  to ground and apply KCL.

   Consider the grounded circuit as shown below.

2622021254_m25.PNG

Applying KCL of point Q we can write incoming current at Q = outgoing current from Q

       $\Rightarrow$      $\frac{V+6}{3}+\frac{V}{1}=\frac{9-V}{5}$

  or      $V\left[\frac{1}{3}+\frac{1}{5}+1\right]=\frac{9}{5}-2$

  or     $V\left[\frac{5+3+15}{15}\right]=\frac{9-10}{5}$

  or        $V\left[\frac{23}{15}\right]=\frac{-1}{5}$

  or      $V=\frac{-3}{23}=-0.13V$

 Thus, current in the 1Ω   resistance  is 0.13 A, from Q to P