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Figure 13.42

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**Figure 13.41** A typical autotransformer. © Todd Systems, Inc. # **Figure 13.42** (a) Step-down autotransformer, (b) step-up autotransformer. (b)

\frac{\mathbf{V}_1}{\mathbf{V}_2} = \frac{N_1}{N_1 + N_2} \tag{13.67}

\frac{\mathbf{I}_1}{\mathbf{I}_2} = \frac{N_1 + N_2}{N_1} = 1 + \frac{N_2}{N_1}

\n(13.68) A major dif ference between con ventional transformers and auto transformers is that the primary and secondary sides of the autotrans former are not only coupled magnetically b ut also coupled conductively. The autotransformer can be used in place of a con ventional transformer when electrical isolation is not required. Example 13.10 Compare the po wer ratings of the tw o-winding transformer in Fig. 13.43(a) and the autotransformer in Fig. 13.43(b). # **Solution:** Although the primary and secondary windings of the autotransformer are together as a continuous winding, they are separated in Fig. 13.43(b) for clarity. We note that the current and voltage of each winding of the autotransformer in Fig. 13.43(b) are the same as those for the two-winding transformer in Fig. 13.43(a). This is the basis of comparing their power ratings. For the two-winding transformer, the power rating is *S*1 = 0.2(240) = 48 VA or *S*2 = 4(12) = 48 VA For the autotransformer, the power rating is *S*1 = 4.2(240) = 1,008 VA or *S*2 = 4(252) = 1,008 VA which is 21 times the power rating of the two-winding transformer. Refer to Fig. 13.43. If the two-winding transformer is a 60-VA, 120 V∕10 V transformer, what is the power rating of the autotransformer? Practice Problem 13.10 **Answer:** 780 VA. Refer to the autotransformer circuit in Fig. 13.44. Calculate: (a) **I**1, **I**2, and **I***o* if **Z***L* = 8 + *j*6 Ω, and (b) the complex power supplied to the load. Example 13.11 # **Solution:** (a) This is a step-up autotransformer with *N*1  =  80, *N*2  =  120, **V**1 = 120⧸ 30°, so Eq. (13.67) can be used to find **V**2 by

\frac{\mathbf{V}_1}{\mathbf{V}_2} = \frac{N_1}{N_1 + N_2} = \frac{80}{200}

or or

\mathbf{V}_2 = \frac{200}{80} \mathbf{V}_1 = \frac{200}{80} (120/30^\circ) = 300/30^\circ \text{ V}

\mathbf{I}_2 = \frac{\mathbf{V}_2}{\mathbf{Z}_L} = \frac{300/30^\circ}{8 + j6} = \frac{300/30^\circ}{10/36.87^\circ} = 30/-6.87^\circ \text{ A}

But But

\frac{\mathbf{I}_1}{\mathbf{I}_2} = \frac{N_1 + N_2}{N_1} = \frac{200}{80}

or or

\mathbf{I}_1 = \frac{200}{80} \mathbf{I}_2 = \frac{200}{80} (30 \angle -6.87^\circ) = 75 \angle -6.87^\circ \text{ A}

Atthetap,KCLgives At the tap, KCL gives

\mathbf{I}_1 + \mathbf{I}_o = \mathbf{I}_2

or or

I_o = I_2 - I_1 = 30 \underline{\textstyle{\frac{1}{6.87^\circ}} - 75 \underline{\textstyle{\frac{1}{6.87^\circ}}}} = 45 \underline{\textstyle{\frac{173.13^\circ}{173.13^\circ}}}

(b)Thecomplexpowersuppliedtotheloadis (b) The complex power supplied to the load is

\mathbf{S}_2 = \mathbf{V}_2 \mathbf{I}_2^* = |\mathbf{I}_2|^2 \mathbf{Z}_L = (30)^2 (10/36.87^\circ) = 9/36.87^\circ \text{ kVA}

# <span id="page-604-0"></span>Practice Problem 13.11 For Practice Prob. 13.11. In the autotransformer circuit of Fig. 13.45, find currents **I**1, **I**2, and **I***o*. Take **V**1 = 8 kV, **V**2 = 2 kV. **Answer:** 2 A, 8 A, 6 A.