Question

Abhishek is trying to solve this circuit question using Kirchhoff’s Voltage law.
And he arrives at this equation for the middle loop.
$(-I_2\times 6000)+(-I_3\times 3000)+(-I_4\times 4000)=0$
Is he correct?

• A. Yes
• B. No

Answer 1

The correct option is B No

Let’s try to apply Kirchhoff’s loop law to the middle loop. First let’s name the junctions.
Have named them A, B, C, D as you can see.


Let’s pick a junction and move around in a loop say ABCDA
(remember $\Delta V=ir)$
On moving from A to B, as we move in the direction of current the potential decreases
so it’s $[-I3\times 3000]$ (following the convention where decrease in potential is negative and rise is positive)
On moving from B to C, as we move in the direction of current again the potential decreases
so it’s $[-I4\times 4000]$
On moving from C to D, as we move in the direction of current the potential should decrease but since R there is zero so the potential difference is also zero.
So it’s $[-I3\times 0]=0$.
On moving from D to A, as we move opposite to the direction of current the potential increases
so it’s $[+I2\times 6000]$
So the equation becomes
$-I3\times 3000-I4\times 4000+0+I2\times 6000=0$
which is not same as the equation given in question ( check once!)
So it’s not correct!