Question

Use source transformation on the circuit shown in Fig 4.98 to find \( I_x \) .

Answer 1

solution

we have 

\( Step1 \)

Transform the given current source as shown dependent Source as shown below. 

\( Step2 \)

Combining the 60\( \Omega \) with the 10\( \Omega \) and transform the dependent 

Source as shown below.

\( Step3 \)

Combining 30\( \Omega \) and 70\( \Omega \) gives,

30||70=70\( \times \)\( \frac{30}{100} \)=21Ohms

\( Step4 \)

Transform the dependent current source as shown below.

\( Step5 \)

Applying the KVL to the loop gives.

45\( I_x \)-12+2.1\( I_x \)=0

=>\( I_x=\frac{12}{47.1} \)=254.8mA

Answer

Therefore, \( I_x \)=254.8mA