# Ideal Autotransformers example problem (step-down configuration)

### For the following circuit, calculate the average power delivered to the load.

All values are RMS

Recall the turns ratio expression for the primary and secondary voltages of a step-down autotransformer: $$\frac{\mathbb{V}_2}{\mathbb{V}_1} = \frac{N_2}{N_1+N_2}$$ For our circuit this gives us: $$\frac{\mathbb{V}_2}{\mathbb{V}_1} = \frac{200}{1200} = \frac{1}{6}$$ V1 in terms of V2 gives us: $$\mathbb{V}_1 = 6\mathbb{V}_2 \qquad,(Eqn\;1)$$ Now, recall the turns ratio expression for the primary and secondary currents of a step-down autotransformer: $$\frac{\mathbb{I}_2}{\mathbb{I}_1} = \frac{N_1+N_2}{N_2}$$ For our circuit this gives us: $$\frac{\mathbb{I}_2}{\mathbb{I}_1} = \frac{1200}{200} = 6$$ I1 in terms of I2 gives us: $$\mathbb{I}_1 = \frac{\mathbb{I}_2}{6} \qquad,(Eqn\;2)$$

### Apply Kirchhoff's Voltage Law (KVL) to primary and secondary side.

For the primary we have: $$\mathbb{I}_1(30+j12)+\mathbb{V}_1 = 440$$ $$32.31\;\mathbb{I}_1 \angle 21.8^{\circ}+\mathbb{V}_1 = 440$$ Substitute equations 1 and 2 into the above expression: $$(32.31 \angle 21.8^{\circ}) \Big( \frac{\mathbb{I}_2}{6} \Big)+6\mathbb{V}_2 = 440$$ $$5.385\;\mathbb{I}_2 \angle 21.8^{\circ} + 6\mathbb{V}_2 = 440 \qquad,(Eqn\;A)$$ Applying KVL to the secondary side gives us: $$\mathbb{I}_2(20-j40) - \mathbb{V}_2 = 0$$ $$44.72 \; \mathbb{I}_2 \angle 296.6^{\circ} - \mathbb{V}_2 = 0 \qquad,(Eqn\;B)$$

### Solve equations A and B.

I won't walk through the steps of solving the equations since the method is the same as in the step-up example problem. Using whatever technique you choose (substitution, Cramer's Rule, matrix row operations, etc...) will give you the following value for current I2: $$\mathbb{I}_2 = 1.638\angle 62.3^{\circ} \; A$$

### Calculate the power delivered to the load.

Converting the load impedance from rectangular to polar form gives us: $$\mathbb{Z}_L = 20-j40 = 44.72\angle 296.6^{\circ} \; \Omega$$ Recognizing that our voltage and current values are RMS, and recalling the expressions for complex power, we realize that we can express the complex power delivered to the load as: $$\mathbb{S} = I_{rms}^2 \mathbb{Z}$$ For our circuit we have: $$\mathbb{S}_L = (I_2)^2 \mathbb{Z}_L = (1.638)^2(44.72\angle 296.6^{\circ})$$ $$\mathbb{S}_L = 120\angle 296.6^{\circ} = 53.72-j107.3$$ The real/average power delivered to the load is defined as: $$P_L = R_e \{ \mathbb{S}_L \} = R_e \{ 53.72-j107.3 \}$$

$$P_L = 53.7 \; W$$