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.

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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

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