Figure for the practice problem 13.9

I am studying Fundamental of Electric Circuits by Alexander and Sadiku, and I am not able to solve a practice problem 13.9 about ideal transformer in Chapter 13 of the textbook. My answer is different from that provided by the textbook and I am not able to find a reference solution on internet.

As shown in the above plot, the problem asks to find \$V_0\$.

I assumes the currents of the three meshes (bottom left, bottom right and top right) are \$I_1, I_2, I_3\$ defined in clockwise directions, and the voltages at the two sides of the ideal transformers are \$V_1, V_2\$ with positive defined at the points marked by red dot. Then I wrote down the following equations

$$120 - 4 I_1 - V_1 = 0 \,\,\,\;\;\;\;\;\;\;(1)$$ $$-V_2 - 2 (I_2 - I_3) - 8 I_2 = 0\;\;\;\;\;\;\;(2)$$ $$-8 I_3 - 2 (I_3 - I_2) = 0\;\;\;\;\;\;(3) $$ $$V_1 = V_2 /2 \;\;\;\;\;\;(4)$$ $$V_1 (I_1 - I_3 ) = -V_2 (I_2 - I_3)\;\;\;\;\;\;(5)$$

Here Eqs.(1-3) are due to Kirchhoff's voltage law (KVL), Eq.(4-5) are due to ideal transformer, while the last one states that there is no energy loss. We note that the currents go through the two sides of the transformer are \$(I_1-I_3)\$ and \$(I_2-I_3)\$ respectively.

However, by solving the above equations, I got \$V_0=-18.46\$ V, different from the result provided by the textbook \$V_0=48\$ V. I checked several times the equations and the solutions, but cannot figure out my mistake.


I used the following short mathematica script to solve the above equations, if it provides any convenience.

 sol = Solve[120 - 4 I1 - V1 == 0 
  && -V2 - 2 (I2 - I3) - 8 I2 == 0 
  && -8 I3 - 2 (I3 - I2) == 0 && V1 ==    V2 /2  
  && V1 (I1 - I3 ) ==  -V2 (I2 - I3) , {I1, I2, I3, V1, V2}]
 N[8 I3 ] /. sol


Bumping my brain against user287001's nodal analysis, I found the mistake in my mesh analysis. The third equation above jumps from one side of the transformer to anther while overlooking the voltage difference. The correct one reads

$$-8 I_3 - 2 (I_3 - I_2) + (V_1 + V_2) = 0\;\;\;\;\;(3') $$

Calculations using Mathematica confirmed the result. Lesson learned.


3 Answers 3


Many of us have quite low brain capacity. For example your equations have too much variables at least for me. I can't at a glance see, is somewhere an error. But here's one proper solution. The facts are collected to equations with smaller number of variables: enter image description here

The equations are

  • Kirchoff's current laws in nodes A and B
  • an expression for the current of R2 (actually it's Kirchoff's voltage law in loop R2-R3-transformer)

All terms are moved to the left side to get zeros to the right sides.

The solution is found by MS Excel Solver:

enter image description here

The orange area has the modified variables. The blue area has the left sides of the three equations. The yellow area is the goal - make this zero. It's the sum of the squares of the left sides.

The result: V0 = 48V

  • \$\begingroup\$ really appreciated! \$\endgroup\$
    – gamebm
    Commented Oct 7, 2017 at 18:40
  • \$\begingroup\$ Nice Excell solver by the way! \$\endgroup\$
    – gamebm
    Commented Oct 7, 2017 at 18:47

The 48v at least in simulation seems plausible (~47.7v):

enter image description here

  • \$\begingroup\$ Thx for the help! \$\endgroup\$
    – gamebm
    Commented Oct 7, 2017 at 18:40
  1. Mesh 1: 4(i1) + V1 = 120
  2. Mesh 2: 10(i2) - 2(i3) - V2 = 0
  3. Mesh 3: 10(i3) - 2(i2) + V2 - V1 = 0

Thus we got the equations for the three loops through mesh analysis.

Now transformation ratio is given.

From that : V2/V1 = 2
(Current in primary / Current in secondary) = 2
(i1-i3)/(i3-i2) = 2

From above we get:
V2 = -2V1 (- due to the polarity)
i2 = (3i3 - i1)/2

Substituting in the three mesh equations, we will get:

  1. 4i1 + V1 = 120
  2. -5i1 + 13i3 + 2V1 = 0
  3. i1 + 7i3 -3V1 = 0

Solving the three equations:

i1 = 24.46 A
i3 = 6 A
V1 = 22.15 V

V0 = ir = 6(8) = 48V


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