# How do you solve the following dependent circuit problem? I tried to use KVL to get i1, but I got .44, which is the wrong answer.

#### mcropper14

Joined Dec 4, 2019
3
Circuit problem:
I have the answers to it, but I need to know the process of how to get there.

#### Jony130

Joined Feb 17, 2009
5,488
You need to show us your calculations.

#### mcropper14

Joined Dec 4, 2019
3
I0 = I1 + I3
Smaller circuit for i1: 11 - 62 -19I1 = 0
i1 = -2.68

i3 = 11 - 15I - 2.68 - 9 = 0
i3 = -.045

#### Jony130

Joined Feb 17, 2009
5,488
Smaller circuit for i1: 11 - 62 -19I1 = 0
i1 = -2.68
Your equation is wrong. And you do not see it because you do not track the units properly.
11 of what?
62 alone without any unit?
19*I1 and the unit is?

#### WBahn

Joined Mar 31, 2012
30,062
I0 = I1 + I3
Smaller circuit for i1: 11 - 62 -19I1 = 0
i1 = -2.68

i3 = 11 - 15I - 2.68 - 9 = 0
i3 = -.045
Several issues.

First, your thread title indicates that you are supposed to use KVL, yet you're working with KCL equations.

Second, your equations use currents that aren't defined in your diagram. What is I0? I have no idea, because you haven't defined it.

Third, you are making mistakes that would be trivially easy to catch if you had tracked your units properly, but because you aren't do so, you don't catch it and then are left wondering why you aren't getting the correct result.

"Smaller circuit for i1: 11 - 62 -19I1 = 0"

should be

(11 V) - (62 Ω) - (19 Ω)(I1) = 0

So you are subtracting a resistance from a voltage. Not physically possible to do that, so you KNOW the equation is wrong right at this point. No need to waste your time going any further until you address this.

#### mcropper14

Joined Dec 4, 2019
3
Several issues.

First, your thread title indicates that you are supposed to use KVL, yet you're working with KCL equations.

Second, your equations use currents that aren't defined in your diagram. What is I0? I have no idea, because you haven't defined it.

Third, you are making mistakes that would be trivially easy to catch if you had tracked your units properly, but because you aren't do so, you don't catch it and then are left wondering why you aren't getting the correct result.

"Smaller circuit for i1: 11 - 62 -19I1 = 0"

should be

(11 V) - (62 Ω) - (19 Ω)(I1) = 0

So you are subtracting a resistance from a voltage. Not physically possible to do that, so you KNOW the equation is wrong right at this point. No need to waste your time going any further until you address this.

So why is it not:
- (62) ohms - (19 ohms) (i1) = 0, I thought that 62 and 19 were voltage drops.

#### WBahn

Joined Mar 31, 2012
30,062
So why is it not:
- (62) ohms - (19 ohms) (i1) = 0, I thought that 62 and 19 were voltage drops.
The "62" is from the resistance of R1, which is 62 Ω. Similar for the "19" and R2.

Resistance is NOT a voltage drop. To get the voltage drop, you need to multiply the resistance by the current flowing through that resistance. Review Ohm's Law.