# Ideal Autotransformers example problem (step-up configuration)

### For the following circuit:

All values are RMS

• A) Calculate I1, I2 and I0.
• B) Determine the average power delivered to the load.

### A) Calculate current values

Recall the turns ratio expression for the primary and secondary voltages of a step-up autotransformer: $$\frac{\mathbb{V}_2}{\mathbb{V}_1} = \frac{N_1+N_2}{N_1}$$ For our example we will get the following: $$\frac{\mathbb{V}_2}{\mathbb{V}_1} = \frac{280}{80} = 3.5$$ If we express V1 in terms of V2 we get: $$\mathbb{V}_1 = \frac{\mathbb{V}_2}{3.5} \qquad ,(Eqn\;1)$$ Now, recall the turns ratio expression for the primary and secondary currents of a step-up autotransformer: $$\frac{\mathbb{I}_2}{\mathbb{I}_1} = \frac{N_1}{N_1+N_2}$$ For our example we have: $$\frac{\mathbb{I}_2}{\mathbb{I}_1} = \frac{80}{280} = \frac{1}{3.5}$$ If we express I1 in terms of I2 we get: $$\mathbb{I}_1 = 3.5\mathbb{I}_2 \qquad ,(Eqn\;2)$$

## Apply Kirchhoff's Voltage Law (KVL) to the primary side:

$$\mathbb{I}_1(2-j6) + \mathbb{V}_1 = 120\angle 45^{\circ}$$ $$6.325\;\mathbb{I}_1 \angle 288.4^{\circ} + \mathbb{V}_1 = 120\angle 45^{\circ}$$ Substitute equations 1 and 2 into the above expression and get the following: $$(6.325 \angle 288.4^{\circ})(3.5\;\mathbb{I}_2)+\frac{\mathbb{V}_2}{3.5} = 120\angle 45^{\circ}$$ $$22.14\;\mathbb{I}_2 \angle 288.4^{\circ} + 0.2857\; \mathbb{V}_2 = 120\angle 45^{\circ} \qquad ,(Eqn\;A)$$

## Apply (KVL) to the secondary side:

$$\mathbb{I}_2(10+j40) - \mathbb{V}_2 = 0$$ $$41.23\;\mathbb{I}_2 \angle 75.96^{\circ} - \mathbb{V}_2 = 0 \qquad ,(Eqn\;B)$$

## Solve equations A and B:

We will use Cramer's Rule to solve for I2 and V2 in equations A and B. First put the equations in matrix form: $$\begin{pmatrix} 22.14\angle 288.4^{\circ}&0.2857&120\angle 45^{\circ}\\ 41.23 \angle 75.96^{\circ}&-1&0\\ \end{pmatrix}$$ Now calculate the required determinants: $$D = \begin{vmatrix} 22.14\angle 288.4^{\circ}&0.2857\\ 41.23 \angle 75.96^{\circ}&-1\\ \end{vmatrix} = 13.74 \angle 135.8^{\circ}$$ $$D_{\mathbb{I}_2} = \begin{vmatrix} 120\angle 45^{\circ}&0.2857\\ 0&-1\\ \end{vmatrix} = 120 \angle 225^{\circ}$$ $$D_{\mathbb{V}_2} = \begin{vmatrix} 22.14\angle 288.4^{\circ}&120\angle 45^{\circ}\\ 41.23 \angle 75.96^{\circ}&0\\ \end{vmatrix} = 4948 \angle 301^{\circ}$$ Now calculate V2 and I2: $$\mathbb{V}_2 = \frac{D_{\mathbb{V}_2}}{D} = \frac{4948 \angle 301^{\circ}}{13.74 \angle 135.8^{\circ}}$$ $$\mathbb{V}_2 = 360.1 \angle 165.2 \; V$$ $$\mathbb{I}_2 = \frac{D_{\mathbb{I}_2}}{D} = \frac{120 \angle 225^{\circ}}{13.74 \angle 135.8^{\circ}}$$

$$\mathbb{I}_2 = 8.734 \angle 89.2^{\circ} \; A$$

## Calculate remaining current values:

Using equation #2 and the value we just determined for I2, calculate I1: $$\mathbb{I}_1 = 3.5\mathbb{I}_2 = 3.5(8.734 \angle 89.2^{\circ})$$

$$\mathbb{I}_1 = 30.57 \angle 89.22^{\circ} \; A$$

Now apply Kirchhoff's Current Law (KCL) at the transformer tap: $$\mathbb{I}_1 + \mathbb{I}_0 = \mathbb{I}_2$$ $$\mathbb{I}_0 = \mathbb{I}_2 - \mathbb{I}_1$$ $$\;\;\; = (8.734 \angle 89.2^{\circ}) - (30.57 \angle 89.22^{\circ})$$

$$\mathbb{I}_0 = 21.84\angle 269.2^{\circ} \; A$$

### Determine the average power delivered to load

Recall from our derived expressions for complex power, that when dealing with RMS values, we can use the following: $$\mathbb{S} = \mathbb{V}_{rms} \mathbb{I}_{rms}*$$ So, for the complex power delivered to the load in our circuit (and remembering that all calculated values have been RMS) we would have: $$\mathbb{S}_L = \mathbb{V}_2 \mathbb{I}_2*$$ $$\;\;\;\; = [360.1 \angle 165.2][8.734 \angle (-89.2^{\circ})]$$ $$\mathbb{S}_L = 3145\angle 76^{\circ} = 760.9+j3052$$ Finally, we know that the average power (or real power) equates to: $$P_L = R_e \{ \mathbb{S}_L \} = R_e \{ 760.9+j3052 \}$$

$$P_L = 760.9 \; W$$

Next we will look at an example problem involving a step-down autotransformer.

Continue on to autotransformers (step-down example problem)...