# Negative voltage in Mesh Current Analysis

I am trying to learn about Mesh Current Analysis and have a question

I was reading this explanation on the subject from electronics-tutorials.ws:

http://www.electronics-tutorials.ws/dccircuits/dcp_5.html

and one of the equations is $-20 = -40I_1 + 60I_2$

I understand where $-40I_1 + 60I_2$ comes from, and I'm assuming -20 is taken from the voltage and it is given as negative because the loop goes clockwise, can someone confirm why -20 is negative, please?

Thanks

Right, the -20 in that equation is negative because the 20 V voltage source would tend to create a current in the opposite direction of (the loop current) I2.

Remember that Mesh Analysis is a systematic application of KVL Law which states that the sum of the voltages in a closed loop is equal to 0, i.e. The voltage across the 40Ω resistor plus the voltage across the 20Ω resistor plus the 20V source has to equal 0... (Voltage Rise = Voltage Drop)

So if we call the 40Ω R3 and the 20Ω R2 we have that...

R3 is a Voltage rise (by passive sign convention), R2 looks like a voltage drop but its actually a rise as well because the direction of i2 is reversed because is flowing anti-clockwise (according to the passive sign convention) and since the current i2 is leaving the positive terminal of the 20V source is a voltage rise as well, now we can set up our equation...

-R3(I2-I1)+R2(-I2)+(-20v) = 0

-40(I2-I1)+20(-I2)+(-20v) = 0

40(I1)-60(I2)-20V = 0

-20V = - 40(I1) + 60(I2)