# Solve By Source Definitions, KCL and KVL

Find the voltage across the current source and the current passing through the voltage source.

Assume that $I_1=3A$ , $R_1=2 \Omega$ , $R_2=3 \Omega$ , $R_3=2 \Omega$ , $I_1=3A$ , $V_1=15 V$ ,

Solution
$R_1$ is in series with the current source; therefore, the same current passing through it as the current source:

$I_{R_1}=I_1=3A$
and the voltage across $R_1$ can be found by Ohm's law:
$V_{R_1}=R_1 \times I_{R_1} =2 \times 3 = 6V$
To find the voltage across the current source, KVL can be applied around the left hand side loop:

The direction does not matter and would not change the result.
$-V_{I_1}+V_{R_1}-V_1=0A$
$V_{I_1}=V_{R_1}-V_1=6-15=-9 V$ .

$R_2$ and $R_3$ are also in series and their equivalent is $R_{23}=R_2+R_3=3 \Omega + 2 \Omega=5 \Omega$

$R_{23}$ is parallel with the voltage source. This means that its voltage is equal to the voltage of the voltage source.
$V_{R_{23}}=V_1=15 V$
Now, using the Ohm's law, the current passing through $R_{23}$ can be calculated:
$I_{R_{23}}=\frac{V_{R_{23}}}{R_{23}}=\frac{15}{5}=3 A$
To find the current of the voltage source, we can apply KCL at one of the nodes:

$I_{R_{23}}+I_1+I_{V_1}=0$

$I_{V_1}=-I_{R_{23}}-I_1=-3-3=-6 A$
Now, you tell me below what the power of each sources are?

## 5 thoughts on “Solve By Source Definitions, KCL and KVL”

1. plz solve exercise problems on Nortons theorem from BE 3 Sem na books

1. Have you checked out ?
And let me know which problem you would like me to solve.

2. ramasubramanian says:

i will need some kvl&kcl simple problem

3. selvasubramaniyan says:

super

4. This site is really for us student....will like to keep in touch