Q58P - Introduction to Electrodynamics

Question

A transformer (Prob. 7.57) takes an input AC voltage of amplitude ${\mathit{V}}_{\mathbf{1}}$, and delivers an output voltage of amplitude ${\mathit{V}}_{\mathbf{2}}$, which is determined by the turns ratio $\left(\frac{V_2}{V_1}=\frac{N_2}{N_1}\right)$. If ${\mathit{N}}_{\mathbf{2}}\mathbf{>}{\mathit{N}}_{\mathbf{1}}$, 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 ${\mathit{M}}^{\mathbf{2}}\mathbf{=}{\mathit{L}}_{\mathbf{1}}{\mathit{L}}_{\mathbf{2}}$, where $\mathit{M}$ is the mutual inductance of the coils, and ${\mathit{L}}_{\mathbf{1}}\mathbf{,}\mathbf{ }{\mathit{L}}_{\mathbf{2}}$, are their individual self-inductances.

(b) Suppose the primary is driven with AC voltage ${\mathit{V}}_{\mathbf{i}\mathbf{n}}\mathbf{=}{\mathit{V}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\left(\omega t\right)$, and the secondary is connected to a resistor, $\mathit{R}$. Show that the two currents satisfy the relations

$$\begin{gathered} {\mathit{L}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{d}{\mathbf{l}}_{\mathbf{1}}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathit{M}\frac{\mathbf{d}{\mathbf{l}}_{\mathbf{2}}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\mathit{V}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\left(\omega t\right)\mathbf{;} \\ {\mathit{L}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{d}{\mathbf{l}}_{\mathbf{2}}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathit{M}\frac{\mathbf{d}{\mathbf{l}}_{\mathbf{1}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}{\mathit{I}}_{\mathbf{2}}\mathit{R}\mathbf{.} \end{gathered}$$

(c) Using the result in (a), solve these equations for ${\mathit{l}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$ and ${\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$. (Assume ${\mathit{l}}_{\mathbf{2}}$ has no DC component.)

(d) Show that the output voltage $\left(V_{out}=l_{2}R\right)$divided by the input voltage $\left(V_{in}\right)$ is equal to the turns ratio: $\frac{{\mathbf{V}}_{\mathbf{o}\mathbf{u}\mathbf{t}}}{{\mathbf{V}}_{\mathbf{i}\mathbf{n}}}\mathbf{=}\frac{{\mathbf{N}}_{\mathbf{2}}}{{\mathbf{N}}_{\mathbf{1}}}$ .

(e) Calculate the input power $(P_{in}=V_{in}{l}_1)$and the output power $\mathbf{(}{\mathbf{P}}_{\mathbf{out}}\mathbf{=}{\mathbf{V}}_{\mathbf{out}}{\mathbf{l}}_{\mathbf{2}}\mathbf{)}$, and show that their averages over a full cycle are equal.

Step-by-Step Solution

Answer

(a) The equation $L_1{L}_2=M^2$ is verified.

(b) The two currents satisfy the relations.

(c) The value of currents is $I_1\left(t\right)=\frac{V_1}{L_1}\left(\frac{1}{\omega }\sin \omega t+\frac{L_2}{R}\cos \omega t\right)$and $I_2\left(t\right)=\frac{-L_2{V}_1\cos \omega t}{MR.}.$

(d) The ratio $\frac{V_{out}}{V_{in}}=-\frac{N_2}{N_1}$is verified.

(e) The input power is $P_{in}=\frac{(V_1{)}^2}{L_1}\left(\frac{1}{\omega }\sin \omega t \cos \omega t+\frac{L_2}{R}{\cos }^2\omega t\right)$, output power is $\frac{(L_2{V}_1{)}^2}{M^{2}R}co{s}^2 \omega t$ and their averages over a full cycle are equal.

Step 1: Determine the mutual inductance

The working of a transformer is based on the ‘mutual inductance’ principal. There are two coils in a transformer, primary and secondary coils. 

 

When the current flows in the primary coil, based on ‘mutual inductance’ it induces a certain emf in the secondary coil which opposes the change in current.

Step 2(a): Relation between the mutual inductance and the individual self-inductances of the coils

Assume, the current flowing in the first and second coils of a transformer are $I_1$ and $I_2$ respectively.

 

Then the formula for the magnetic flux through the first coil is given by,

$$\begin{gathered} {\Phi }_1=I_1{L}_1+M{I}_2 \\ {N}_1\Phi =I_1{L}_1+M{I}_2 \\ \Phi =I_1\frac{L_1}{N_1}+I_2\frac{M}{N_1} \end{gathered}$$

 

Here, $\mathrm{\Phi }$ is the flux through one turn of the coil.

 

Similarly, the formula for the magnetic flux through the second coil is given by,

 

$$\begin{gathered} {\Phi }_2=I_2{L}_2+M{I}_1 \\ {N}_2\Phi =I_2{L}_2+M{I}_1 \\ \Phi =I_2\frac{L_2}{N_2}+I_1\frac{M}{N_2} \end{gathered}$$

 

Combining both expressions,

$I_1\frac{L_1}{N_1}+I_2\frac{M}{N_1}=I_2\frac{L_2}{N_2}+I_1\frac{M}{N_2}$

If $I_1=0$, then,

$$\begin{gathered} 0+I_2\frac{M}{N_1}=I_2\frac{L_2}{N_2}+0 \\ \frac{M}{N_1}=\frac{L_2}{N_2} \end{gathered}$$

If $I_2=0$, then,

$$\begin{gathered} I_1\frac{L_1}{N_1}+0=0+I_1\frac{M}{N_2} \\ \frac{L_1}{N_1}=\frac{M}{N_2} \\ \frac{M}{L_1}=\frac{L_2}{M} \\ {L}_1{L}_2=M^2 \end{gathered}$$

Hence, the equation is verified.

Step 3(b): Verify the two current relations

The formula for the emf induced in the first coil is given by,

 

$$\begin{gathered} -{\epsilon }_0=\frac{d{\Phi }_1}{dt} \\ -{\epsilon }_1=L_1\frac{d{l}_1}{dt}+M\frac{d{l}_2}{dt} \\ -{\epsilon }_1=V_1\cos \left(\omega t\right) \end{gathered}$$

 

Similarly, the formula for the emf induced in the first coil is given by,

$$\begin{gathered} -{\mathrm{\epsilon }}_2=\frac{{\mathrm{d\Phi }}_2}{\mathrm{dt}} \\ -{\epsilon }_2=L_2\frac{d{l}_2}{dt}+M\frac{d{l}_1}{dt} \\ -{\epsilon }_2=-l_{2}R \end{gathered}$$

 

Hence proved, the two currents satisfy the relations.

Step 4(c): Solving equations for current values

The first relation is given as,

 

$L_1\frac{d{I}_1}{dt}+M\frac{d{I}_2}{dt}$

 

Multiplying both sides by $L_2$ ,

 

$L_1{L}_2\frac{d{I}_1}{dt}+L_2\frac{d{I}_2}{dt}M=L_2{V}_{1}cos \omega t$

 

Substitute, $L_2\frac{d{I}_2}{dt}=-I_{2}R-M\frac{d{I}_1}{dt}.$

$$\begin{gathered} L_1{L}_2\frac{d{I}_1}{dt}+\left(-I_{2}R-M\frac{\left(d{I}_1\right)}{dt}M\right)=L_2{V}_{1}cos\omega t \\ {M}^2\frac{d{I}_1}{dt}-MR{I}_2-M^2\frac{d{I}_2}{dt}=L_2{V}_{1}cos \omega t \\ {I}_2\left(t\right)=\frac{-L_2{V}_1\cos \omega t}{MR} \end{gathered}$$

Differentiating both sides w.r.t. t.

$$\begin{gathered} \frac{d{l}_2}{dt}=\frac{d}{dt}\left(\frac{-L^2{V}_1\cos \omega t}{MR}\right) \\ \frac{d{l}_2}{dt}=\frac{L_2{V}_1\cos \omega t}{MR} \end{gathered}$$

Substitute the value of $\frac{d{l}_2}{dt}$in first relation,

$$\begin{gathered} L_1\frac{d{I}_1}{dt}+M\frac{L_2{V}_1\omega \sin \omega t}{MR}=V_1 cos \omega t \\ {L}_1\frac{d{I}_1}{dt}=V_{1}cos \omega t-M\left(\frac{L_2{V}_1 \omega \sin \omega t}{MR}\right) \\ \frac{d{I}_1}{dt}=\frac{V_1}{L_1}\left(\cos \omega t-\frac{L_2}{r}\omega \sin \omega t\right) \end{gathered}$$

Integrating both sides,

 

$I_1\left(t\right)=\frac{V_1}{L_1}\left(\frac{1}{\omega }\sin \omega t+\frac{L_2}{R} \cos \omega t\right)$

 

Hence, the equation for currents $I_1\left(t\right)$ and $I_2 \left(t\right)$ is explained.

Step 5(d): The ratio of the output voltage to the input voltage

The formula for the ratio of the output voltage to the input voltage is given by,

$\frac{V_{out}}{V_{in}}=\frac{I_{2}R}{V_{1 }cos \omega t}$

Substitute value of $I_2$.

 

$$\begin{gathered} \frac{V_{out}}{V_{in}}=\frac{{\displaystyle \frac{-L_2{V}_1\cos \omega t\times R}{MR}}}{V_1\cos \omega t} \\ \frac{V_{out}}{V_{in}}=\frac{-L_2}{M} \\ \frac{V_{out}}{V_{in}}=\frac{N_2}{N_1} \end{gathered}$$

 

Hence, the ratio of the output voltage to the input voltage is $-\frac{N_2}{N_1}$ .

Step 6(e): The input power, output power and their averages over a full cycle

The formula for the input power to the transformer is given by,

 

$$\begin{gathered} P_{in}=V_n{I}_1 \\ {P}_{in}=\left(V_1\cos \omega t\right)\left(\frac{V_1}{L_1}\right)\left(\frac{1}{\omega }\sin \omega t+\frac{L_2}{R}\cos \omega t\right) \\ {P}_{in}=\frac{{\left(V_1\right)}^2}{L_1}\left(\frac{1}{\omega }\sin \omega t \cos \omega t+\frac{L_2}{R}{\cos }^2\omega t\right) \end{gathered}$$

 

In the formula for the average power input ${\left(P_{in}\right)}_{avg}$ over the full cycle, Substitute the average value of ${\cos }^2 \omega t=\frac{1}{2}$ and $\sin \omega t \cos \omega t=0$ in expression,

 

$$\begin{gathered} (P_{in}{)}_{avg}=\frac{(V_1{)}^2}{L_1}\left(\frac{1}{\omega }\times 0+\frac{L_2}{R}\frac{1}{2}\right) \\ (P_{in}{)}_{avg}=\frac{1}{2}(V_1{)}^2\left(\frac{L_2}{L_{1}R}\right) \end{gathered}$$

 

And, the formula for the output power of the transformer is given by,

$$\begin{gathered} P_{out}=V_{out}{I}_2 \\ {P}_{out}=\left(I_2\right)R \\ {P}_{out}=\frac{\left(L_2{V}_1\right)}{M^{2}R}{\cos }^2 \omega t \end{gathered}$$

Similarly, in the formula for the average ${\left(P_{out}\right)}_{avg}$ power output over the full cycle, Substitute the average value of ${\cos }^2\omega t=\frac{1}{2}$ and $M^2=L_1{L}_2$ in expression.

$$\begin{gathered} {\left(P_{out}\right)}_{avg}=\frac{{\left(L_2{V}_1\right)}^2}{L_1{L}_{2}R}\times \frac{1}{2} \\ {\left(P_{out}\right)}_{avg}=\frac{1}{2}{\left(V_1\right)}^2\left(\frac{{\left(l_2\right)}^2}{L_1{L}_{2}R}\right) \\ {\left(P_{out}\right)}_{avg}=\frac{1}{2}{\left(V_1\right)}^2\left(\frac{L_2}{L_{1}R}\right) \end{gathered}$$

Then comparing the average power input and output equations.

 

${\left(P_{out}\right)}_{avg}-{\left(P_{out}\right)}_{avg}-\frac{{\left(V_1\right)}^2{L}_2}{2L_{1}R}$

 

Hence proved, the average power over a full cycle is equal.