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Question 7.58: A transformer (Prob. 7.57) takes an input AC voltage of ampl...

A transformer (Prob. 7.57) takes an input AC voltage of amplitude V_{1} , and delivers an output voltage of amplitude V_{2} , which is determined by the turns ratio \left(V_{2} / V_{1}=N_{2} / N_{1}\right) . \text { If } N_{2}>N_{1}, the output voltage is greater than the input voltage. Why doesn’t this violate conservation of energy? Answer: Power is the product of voltage and current; if the voltage goes up, the current must come down. The purpose of this problem is to see exactly how this works out, in a simplified model.

(a) In an ideal transformer, the same flux passes through all turns of the primary and of the secondary. Show that in this case M^{2}=L_{1} L_{2} , where M is the mutual inductance of the coils, and L_{1}, L_{2} are their individual self-inductances.

(b) Suppose the primary is driven with AC voltage V_{ in }=V_{1} \cos (\omega t), and the secondary is connected to a resistor, R. Show that the two currents satisfy the relations

L_{1} \frac{d I_{1}}{d t}+M \frac{d I_{2}}{d t}=V_{1} \cos (\omega t) ; \quad L_{2} \frac{d I_{2}}{d t}+M \frac{d I_{1}}{d t}=-I_{2} R .

(c) Using the result in (a), solve these equations for I_{1}(t) \text { and } I_{2}(t). (Assume I_{1} has no DC component.)

(d) Show that the output voltage \left(V_{\text {out }}=I_{2} R\right) divided by the input voltage \left(V_{\text {in }}\right) is equal to the turns ratio: V_{\text {out }} / V_{\text {in }}=N_{2} / N_{1} .

(e) Calculate the input power\left(P_{\text {in }}=V_{\text {in }} I_{1}\right) and the output power \left(P_{\text {out }}=V_{\text {out }} I_{2}\right), and show that their averages over a full cycle are equal.

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\text { (a) Suppose current } I_{1} \text { flows in coil } 1 \text {, and } I_{2} \text { in coil } 2 \text {. Then (if } \Phi \text { is the flux through one turn) }:

\Phi_{1}=I_{1} L_{1}+M I_{2}=N_{1} \Phi ; \quad \Phi_{2}=I_{2} L_{2}+M I_{1}=N_{2} \Phi, \quad \text { or } \Phi=I_{1} \frac{L_{1}}{N_{1}}+I_{2} \frac{M}{N_{1}}=I_{2} \frac{L_{2}}{N_{2}}+I_{1} \frac{M}{N_{2}} .

\text { In case } I_{1}=0, \text { we have } \frac{M}{N_{1}}=\frac{L_{2}}{N_{2}} ; \text { if } I_{2}=0 \text {, we have } \frac{L_{1}}{N_{1}}=\frac{M}{N_{2}} . \text { Dividing: } \frac{M}{L_{1}}=\frac{L_{2}}{M}, \text { or } L_{1} L_{2}=M^{2} . qed

\text { (b) }- ε _{1}=\frac{d \Phi_{1}}{d t}=L_{1} \frac{d I_{1}}{d t}+M \frac{d I_{2}}{d t}=V_{1} \cos (\omega t) ;- ε _{2}=\frac{d \Phi_{2}}{d t}=L_{2} \frac{d I_{2}}{d t}+M \frac{d I_{1}}{d t}=-I_{2} R . qed

\text { (c) Multiply the first equation by } L_{2}: L_{1} L_{2} \frac{d I_{1}}{d t}+L_{2} \frac{d I_{2}}{d t} M=L_{2} V_{1} \cos \omega t \text {. Plug in } L_{2} \frac{d I_{2}}{d t}=-I_{2} R-M \frac{d I_{1}}{d t} .

M^{2} \frac{d I_{1}}{d t}-M R I_{2}-M^{2} \frac{d I_{1}}{d t}=L_{2} V_{1} \cos \omega t \Rightarrow I_{2}(t)=-\frac{L_{2} V_{1}}{M R} \cos \omega t . L_{1} \frac{d I_{1}}{d t}+M\left(\frac{L_{2} V_{1}}{M R} \omega \sin \omega t\right)=V_{1} \cos \omega t .

\frac{d I_{1}}{d t}=\frac{V_{1}}{L_{1}}\left(\cos \omega t-\frac{L_{2}}{R} \omega \sin \omega t\right) \Rightarrow I_{1}(t)=\frac{V_{1}}{L_{1}}\left(\frac{1}{\omega} \sin \omega t+\frac{L_{2}}{R} \cos \omega t\right) .

\text { (d) } \frac{V_{\text {out }}}{V_{\text {in }}}=\frac{I_{2} R}{V_{1} \cos \omega t}=\frac{-\frac{L_{2} V_{1}}{M R} \cos \omega t R}{V_{1} \cos \omega t}=-\frac{L_{2}}{M}=-\frac{N_{2}}{N_{1}} . \text { The ratio of the amplitudes is } \frac{N_{2}}{N_{1}} . qed

\text { (e) } P_{\text {in }}=V_{\text {in }} I_{1}=\left(V_{1} \cos \omega t\right)\left(\frac{V_{1}}{L_{1}}\right)\left(\frac{1}{\omega} \sin \omega t+\frac{L_{2}}{R} \cos \omega t\right)=\frac{\left(V_{1}\right)^{2}}{L_{1}}\left(\frac{1}{\omega} \sin \omega t \cos \omega t+\frac{L_{2}}{R} \cos ^{2} \omega t\right).

P_{ out }=V_{ out } I_{2}=\left(I_{2}\right)^{2} R=\frac{\left(L_{2} V_{1}\right)^{2}}{M^{2} R} \cos ^{2} \omega t . \text { Average of } \cos ^{2} \omega t \text { is } 1 / 2 \text {; average of } \sin \omega t \cos \omega t \text { is zero. } .

\text { So }\left\langle P_{\text {in }}\right\rangle=\frac{1}{2}\left(V_{1}\right)^{2}\left(\frac{L_{2}}{L_{1} R}\right) ;\left\langle P_{\text {out }}\right\rangle=\frac{1}{2}\left(V_{1}\right)^{2}\left[\frac{\left(L_{2}\right)^{2}}{M^{2} R}\right]=\frac{1}{2}\left(V_{1}\right)^{2}\left[\frac{\left(L_{2}\right)^{2}}{L_{1} L_{2} R}\right] ; \quad\left\langle P_{\text {in }}\right\rangle=\left\langle P_{\text {out }}\right\rangle=\frac{\left(V_{1}\right)^{2} L_{2}}{2 L_{1} R}.

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