# Reflected Impedance and Power

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In the previous section we found that
$${ E_p \over E_s} = {N_p \over N_s}$$
and
$${ I_p \over I_s} = {N_s \over N_p} = { 1 \over a}$$
Dividing the first by the second, we have
$${ E_p / E_s \over I_p / I_s} = {a \over 1/a}$$ $${ E_p \over I_p }{ I_s \over E_s} =a^2$$ $${ E_p \over I_p } = a^2 { E_s \over I_s }$$
However, since
$$Z_p = { E_p \over I_p }$$
and
$$Z_L = { E_s \over I_s }$$
then
$$Z_p = a^2 Z_L$$
which in words states that the impedance of the primary circuit of an ideal transformer is the transformation ratio squared times the impedance of the load. If a transformer is used, therefore, an impedance can be made to appear larger or smaller at the primary by placing it in the secondary of a step-down (a > 1) or step-up (a < 1) transformer, respectively.
Note that if the load is capacitive or inductive, the reflected impedance will also be capacitive or inductive.
For the ideal iron-core transformer,
$${ E_p \over E_s } = a = {I_s \over I_p}$$
or
$$\bbox[10px,border:1px solid grey]{E_p I_p = E_s I_s} \tag{2}$$
and
$$P_{in} = P_{out}$$
Example 1: For the iron-core transformer of Fig. 1:
a. Find the magnitude of the current in the primary and the impressed voltage across the primary.
b. Find the input resistance of the transformer.
Fig. 1: Example 1.
Solution:
a. ${ I_p \over I_s} = {N_s \over N_p}$
$$I_p = {N_s \over N_p} I_s = ( {5t \over 40t})(0.1 A) = 12.5mA$$ $$V_L = I_sZ_L = (0.1A)(2 kΩ) = 200V$$
Also,
$$E_p = {N_P \over N_S} V_L = ( {40t \over 5t})(200 V) = 1600V$$
b.$Z_p = a^2Z_L$
$$a = {N_p \over N_s} = 8$$ $$Z_p = 8^2 (2 kΩ) = R_p = 128kΩ$$
Example 1: For the residential supply appearing in Fig. 2, determine (assuming a totally resistive load) the following:
a. the value of $R$ to ensure a balanced load
b. the magnitude of $I_1$ and $I_2$
c. the line voltage $V_L$
d. the total power delivered
e. the turns ratio $a = Np /Ns$
Fig. 2: Single-phase residential supply.
Solution:
a.
$$P_{T}=(10)(60 \mathrm{~W})+400 \mathrm{~W}+2000 \mathrm{~W}\\ P_{T}=600 \mathrm{~W}+400 \mathrm{~W}+2000 \mathrm{~W}=3000 \mathrm{~W}\\ P_{in }=P_{out } \text{(purely resistive load)}\\ V_{p} I_{p}=V_{s} I_{s}=3000 \mathrm{~W} \\ (2400 \mathrm{~V}) I_{p}=3000 \mathrm{~W}$$
and
$$I_{p}=1.25 \mathrm{~A}\\ R=\frac{V_{\phi}}{I_{p}}=\frac{2400 \mathrm{~V}}{1.25 \mathrm{~A}}=1920 \Omega$$
b.
$$P_{1}=600 \mathrm{~W}=V I_{1}=(120 \mathrm{~V}) I_{1}$$
and
$$I_{1}=5A\\ P_{2}=2000 \mathrm{~W}=V I_{2}=(240 \mathrm{~V}) I_{2}$$

and
$$I_{2}=8.33 \mathrm{~A}$$

c.
$$V_{L}=\sqrt{3} V_{\phi}=1.73(2400 \mathrm{~V})=\mathbf{415 2} \mathrm{V}$$

d.
$$P_{T}=3 P_{\phi}=3(3000 \mathrm{~W})=9 kW$$

e.
$$a=\frac{N_{p}}{N_{s}}=\frac{V_{p}}{V_{s}}=\frac{2400 \mathrm{~V}}{240 \mathrm{~V}}=10$$

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