Question

V1 R1 + Let I, be the 2 R4 current through the voltage source V1 from the positive to the negative terminal. Find the most efficient way to calculate I, in terms of all the other variables, 11, V1, 12, R1, R3, and R4. Upload Choose a File

Answer 1

The given circuit is:

Solution

Ru R1 يم NA ألم R Iy

It is given that the current I_x   in the voltage source V₁ is flowing from +ve terminal to -ve terminal .

The most efficient way to  find the current $\large I_x$ , is the Source Transformation followed by KVL .

Source Transformation Theorem:

1. A current source in parallel with a resistor can be replaced by a voltage source in series with the same resistor such that the value of voltage source is equal to the product of the source current and the resistor . see the figure below:   

M R I R DIR=V +

2 . Similarly, a voltage source in series with a resistor can be replaced by a current source in parallel with the same resistance. See the figure below:

Here we'll use the 1 st transformation.

>I ㅋ Q. +

• That is, we'll replace the current source  $I_1$ in parallel with R₃ by a Voltage source

$\large V=I_1R_3$ in series with R₃ .  

• Similarly , the current source   
  12
  in parallel with R₄ will be replaced by a voltage source of $\large V=I_2R_4$

in series with R₄ .

NOTE: The polarity of the voltage sources is determined by the arrow head of the current source .The arrow head gives the positive (+) terminal .

So, after applying Source transformation in the two current sources we obtain   the following circuit :

In VĄ R1 M + 1 Ru R3 1. Ru + I R3

Now applying KVL in the above circuit;

$\large I_1R_3-V_1+I_2R_4=I_x(R_3+R_1+R_4)$

$\large =\gg {\color{Blue} I_x=\frac{I_1R_3-V_1+I_2R_4}{R_1+R_3+R_4}}{\color{Red} ...........Answer}$