# Finding voltage drop on a resistance using mesh analysis

How to find out Vx on the specified resistance using mesh analysis?

simulate this circuit – Schematic created using CircuitLab

Here Vx is to be found out on the 1 ohm resistor.

I tried solving it by applying super mesh on the 5A source.I got 3 equations corresponding to the loops.

10*i1 - 11*i2 + 22*i3 - 5 = 0 ----- belongs to the loop containing 5V source
10*i1 - 16*i2 + i3 = 0 ----- belongs to the loop that doesn't have any source
i1 - i3 + 5 = 0 ---- supermesh


The solutions that I got:

i1 ~ 3.28A
i2 ~ 1.943A
i3 ~ -1.718A
Vx ~ 3.661V


But the solutions specified are:

i1 ~ 3.825A
i2 ~ -1A
i3 ~ 0.375A
Vx ~ 1.375A


Can somebody please tell me where did I go wrong?

• Where is polarity defined for Vx? If we assume -ve is the right side , what is that voltage? or using mesh current from 5A Then once Ix is known Vx is known. Dec 13, 2017 at 2:28
• @TonyStewart.EEsince'75 The polarity is + on the left end of 1 ohm and - on the right end of 1 ohm. Dec 13, 2017 at 2:42
• as I expected... en.wikipedia.org/wiki/Mesh_analysis#Supermesh show your loop directions and designations on schematic Dec 13, 2017 at 2:44
• . There are 5 loops to choose from. Dec 13, 2017 at 2:50
• @TonyStewart.EEsince'75 we need to work on 3 of the 5 loops. Right? I worked on the (5V source - 10 ohm- 1 ohm-4 ohm-17ohm) loop, (10 ohm -5 ohm - 1 ohm) loop and the current source calculation. Dec 13, 2017 at 2:57

simulate this circuit – Schematic created using CircuitLab

Assume clockwise direction for all currents in the schematic.

Using KCL: $$I_2 + 5 = I_4$$

For the meshes we get:

10*I2 - 16*I3 +  1*I4 =  0    -----     Loop with no sources
10*I2 - 11*I3 + 22*I4 =  5    -----     Loop with 5V source
1*I2 +  0*I3 -  1*I4 = -5    -----     From previous equation


The equations exactly match with the ones you got. On solving, I got:

I2 ~ -4.156A
I3 ~ -2.545A
I4 ~ +0.844A
Vx ~ +3.389V


All these values are correct as you can test on any simulator.