Q58P A transformer (Prob. 7.57) takes... [FREE SOLUTION]

Chapter 7: Q58P (page 350)

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 $(\frac{V_{2}}{V_{1}} = \frac{N_{2}}{N_{1}})$ . 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 $V_{i n} = V_{1} c o s (蝇 t)$ , and the secondary is connected to a resistor, $R$ . Show that the two currents satisfy the relations
$L_{1} = \frac{d l_{1}}{d t} + M \frac{d l_{2}}{d t} = V_{1} c o s (蝇 t); L_{1} = \frac{d l_{2}}{d t} + M \frac{d l_{1}}{d t} = - I_{2} R .$
(c) Using the result in (a), solve these equations for localid="1658292112247" $l_{1} (t)$ and $l_{2} (t)$ . (Assume $l_{2}$ has no DC component.)
(d) Show that the output voltage $(V_{o u t} = l_{2} R)$ divided by the input voltage $(V_{i n})$ is equal to the turns ratio: $\frac{V_{o u t}}{V_{i n}} = \frac{N_{2}}{N_{1}}$ .
(e) Calculate the input power localid="1658292395855" $(P_{i n} = V_{i n} l_{1})$ and the output power $(P_{out} = V_{out} l_{2})$ , and show that their averages over a full cycle are equal.

Short Answer

Expert verified

(a) The equation $L_{1} L_{2} = M^{2}$ is verified.

(b)The two currents satisfy the relations.

(c) The value of currents is $I_{1} (t) = \frac{V_{1}}{L_{1}} (\frac{1}{蝇} \sin 蝇 t + \frac{L_{2}}{R} \cos 蝇 t)$ and $I_{2} (t) = \frac{- L_{2} V_{1} \cos 蝇 t}{M R .} .$

(d) The ratio $\frac{V_{o u t}}{V_{i n}} = - \frac{N_{2}}{N_{1}}$ is verified.

(e) The input power is $P_{i n} = \frac{(V_{1})^{2}}{L_{1}} (\frac{1}{蝇} \sin 蝇 t \cos 蝇 t + \frac{L_{2}}{R} \cos^{2} 蝇 t)$ , output power is $\frac{(L_{2} V_{1})^{2}}{M^{2} R} c o s^{2} 蝇 t$ and their averages over a full cycle are equal.

Step by step solution

Determine the mutual inductance

The working of a transformer is based on the 鈥榤utual inductance鈥� principal. There are two coils in a transformer, primary and secondary coils.

When the current flows in the primary coil, based on 鈥榤utual inductance鈥� 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,

$桅_{1} = I_{1} L_{1} + M I_{2} N_{1} 桅 = I_{1} L_{1} + M I_{2} 桅 = I_{1} \frac{L_{1}}{N_{1}} + I_{2} \frac{M}{N_{1}}$

Here,role="math" localid="1658293363987" $桅$ is the flux through one turn of the coil.

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

$桅_{2} = I_{2} L_{2} + M I_{1} N_{2} 桅 = I_{2} L_{2} + M I_{1} 桅 = I_{2} \frac{L_{2}}{N_{2}} + I_{1} \frac{M}{N_{2}}$

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,

$0 + I_{2} \frac{M}{N_{1}} = I_{2} \frac{L_{2}}{N_{2}} + 0 \frac{M}{N_{1}} = \frac{L_{2}}{N_{2}}$

If $I_{2} = 0$ , then,

$I_{1} \frac{L_{1}}{N_{1}} + 0 = 0 + I_{1} \frac{M}{N_{2}} \frac{L_{1}}{N_{1}} = \frac{M}{N_{2}} \frac{M}{L_{1}} = \frac{L_{2}}{M} L_{1} L_{2} = M^{2}$

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,

$- 蔚_{0} = \frac{d 桅_{1}}{d t} - 蔚_{1} = L_{1} \frac{d l_{1}}{d t} + M \frac{d l_{2}}{d t} - 蔚_{1} = V_{1} \cos (蝇 t)$

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

role="math" localid="1658294017369" $- 蔚_{2} = \frac{{诲桅}_{2}}{dt} - 蔚_{2} = L_{2} \frac{d l_{2}}{d t} + M \frac{d l_{1}}{d t} - 蔚_{2} = - l_{2} R$

Hence proved, the two currents satisfy the relations.

Step 4(c): Solving equations for current values

The first relation is given as,

$L_{1} \frac{d I_{1}}{d t} + M \frac{d I_{2}}{d t}$

Multiplying both sides by $L_{2}$ ,

$L_{1} L_{2} \frac{d I_{1}}{d t} + L_{2} \frac{d I_{2}}{d t} M = L_{2} V_{1} c o s 蝇 t$

Substitute, $L_{2} \frac{d I_{2}}{d t} = - I_{2} R - M \frac{d I_{1}}{d t} .$

$L_{1} L_{2} \frac{d I_{1}}{d t} + (- I_{2} R - M \frac{(d I_{1})}{d t} M) = L_{2} V_{1} c o s 蝇 t M^{2} \frac{d I_{1}}{d t} - M R I_{2} - M^{2} \frac{d I_{2}}{d t} = L_{2} V_{1} c o s 蝇 t I_{2} (t) = \frac{- L_{2} V_{1} \cos 蝇 t}{M R}$

Differentiating both sides w.r.t. t.

$\frac{d l_{2}}{d t} = \frac{d}{d t} (\frac{- L^{2} V_{1} \cos 蝇 t}{M R}) \frac{d l_{2}}{d t} = \frac{L_{2} V_{1} \cos 蝇 t}{M R}$

Substitute the value of $\frac{d l_{2}}{d t}$ in first relation,

$L_{1} \frac{d I_{1}}{d t} + M \frac{L_{2} V_{1} 蝇 \sin 蝇 t}{M R} = V_{1} c o s 蝇 t L_{1} \frac{d I_{1}}{d t} = V_{1} c o s 蝇 t - M (\frac{L_{2} V_{1} 蝇 \sin 蝇 t}{M R}) \frac{d I_{1}}{d t} = \frac{V_{1}}{L_{1}} (\cos 蝇 t - \frac{L_{2}}{r} 蝇 \sin 蝇 t)$

Integrating both sides,

$I_{1} (t) = \frac{V_{1}}{L_{1}} (\frac{1}{蝇} \sin 蝇 t + \frac{L_{2}}{R} \cos 蝇 t)$

Hence, the equation for currents $I_{1} (t)$ and $I_{2} (t)$ 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,

$\frac{V_{o u t}}{V_{i n}} = \frac{I_{2} R}{V_{1} c o s 蝇 t}$

Substitute value of $I_{2}$ .

$\frac{V_{o u t}}{V_{i n}} = \frac{\frac{- L_{2} V_{1} \cos 蝇 t 脳 R}{M R}}{V_{1} \cos 蝇 t} \frac{V_{o u t}}{V_{i n}} = \frac{- L_{2}}{M} \frac{V_{o u t}}{V_{i n}} = \frac{N_{2}}{N_{1}}$

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,

$P_{i n} = V_{n} I_{1} P_{i n} = (V_{1} \cos 蝇 t) (\frac{V_{1}}{L_{1}}) (\frac{1}{蝇} \sin 蝇 t + \frac{L_{2}}{R} \cos 蝇 t) P_{i n} = \frac{{(V_{1})}^{2}}{L_{1}} (\frac{1}{蝇} \sin 蝇 t \cos 蝇 t + \frac{L_{2}}{R} \cos^{2} 蝇 t)$

In the formula for the average power input ${(P_{i n})}_{a v g}$ over the full cycle, Substitute the average value of $\cos^{2} 蝇 t = \frac{1}{2}$ and $\sin 蝇 t \cos 蝇 t = 0$ in expression,

$(P_{i n})_{a v g} = \frac{(V_{1})^{2}}{L_{1}} (\frac{1}{蝇} 脳 0 + \frac{L_{2}}{R} \frac{1}{2}) (P_{i n})_{a v g} = \frac{1}{2} (V_{1})^{2} (\frac{L_{2}}{L_{1} R})$

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

$P_{o u t} = V_{o u t} I_{2} P_{o u t} = (I_{2}) R P_{o u t} = \frac{(L_{2} V_{1})}{M^{2} R} \cos^{2} 蝇 t$

Similarly, in the formula for the average ${(P_{o u t})}_{a v g}$ power output over the full cycle, Substitute the average value of $\cos^{2} 蝇 t = \frac{1}{2}$ and $M^{2} = L_{1} L_{2}$ in expression.

${(P_{o u t})}_{a v g} = \frac{{(L_{2} V_{1})}^{2}}{L_{1} L_{2} R} 脳 \frac{1}{2} {(P_{o u t})}_{a v g} = \frac{1}{2} {(V_{1})}^{2} (\frac{{(l_{2})}^{2}}{L_{1} L_{2} R}) {(P_{o u t})}_{a v g} = \frac{1}{2} {(V_{1})}^{2} (\frac{L_{2}}{L_{1} R})$

Then comparing the average power input and output equations.

${(P_{o u t})}_{a v g} - {(P_{o u t})}_{a v g} - \frac{{(V_{1})}^{2} L_{2}}{2 L_{1} R}$

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

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

91影视

Short Answer

Step by step solution

Determine the mutual inductance

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

Step 3(b): Verify the two current relations

Step 4(c): Solving equations for current values

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

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Fields in Physics

Engineering Physics

Fundamentals of Physics

Dynamics

Geometrical and Physical Optics

Linear Momentum

Study anywhere. Anytime. Across all devices.