What is the power dissipated in the primary of the ideal transformer in the given circuit ?

Option 4 : 62.5 W

__Concept:__

Power in an electric circuit is a combination of the Voltage and Current in a circuit.

Power in an electrical circuit is defined as:

P = V × I

\(P = {I^2}R = \frac{{{V^2}}}{R}\)

Turns ratio in the transformer

It is defined as the number of turns on its secondary divided by the number of turns in the primary coil.

\(\frac{{{N_2}}}{{{N_1}}} = \frac{{{V_2}}}{{{V_1}}} = \frac{{{I_1}}}{{{I_2}}}\)

N2: Turns in the secondary coil

N1: Turns in the primary coil

V2: Secondary voltage

V1: Primary voltage

I1: Primary current

I2: Secondary current

__Calculation:__

Consider the circuit as shown:

Based on the turns ratio formula the voltage at the secondary winding is:

V2 = 5 × 50

V2 = 250 V

Secondary current is:

\({I_2} = \frac{{250}}{{1000}} = 0.25A\)

Primary current is:

\({I_1} = {I_2}\frac{{{N_2}}}{{{N_1}}}\)

\({I_1} = 0.25 \times \frac{5}{1}\)

I1 = 1.25 A

Power dissipated in the primary is:

P = 50 × 1.25

P = 62.5 W

**Alternative method:**

Referring secondary load to the primary

\(R_2' = \frac{R_2}{K^2}\)

K: Turns ratio

\(R_2' = \frac{1k\Omega}{25} = 40 \Omega\)

I1 = 50∠0° / 40

I1 = 1.25 A

P = I12 × R

P = 1.252 × 40

P = 62.5 W

Option 1 : coper loss = iron loss

Copper loss:

A loss in a transformer that takes place in the winding resistance of a transformer is known as copper loss. These are variable losses.

Copper loss = i2R

Iron loss:

Iron losses = Eddy current loss + hysteresis loss

Eddy current loss = Ket2f2B2

Hysteresis loss = KhfBx

These are constant losses.

Efficiency:

The efficiency of the Transformer \((\eta ) = \frac{{XS\cos \phi }}{{XS\cos \phi + {P_i} + {X^2}{P_e}}}\)

Where, X = Fraction of load

S = Apparent power in kVA

Wi = Iron losses

Wcu = Copper losses

The transformer will give the maximum efficiency when their copper loss is equal to the iron loss.

Efficiency is maximum at some fraction x of full load.

\(x = \sqrt {\frac{{{W_i}}}{{{W_{cu}}}}}\)

Where Wi = iron losses

Wcu = copper losses

kVA at maximum efficiency is given by,

\(kVA\;at\;{\eta _{max}} = full\;load\;kVA \times \sqrt {\frac{{{W_i}}}{{{W_{cu}}}}}\)

If an ideal transformer has an inductive load element at port 2 as shown in the figure below, the equivalent inductance at port 1 is

Option 2 : \({n^2}L\)

__ Concept__:

Transformation ratio of transformer is given by K = V2/V1 = E2/E1 = N2/N1.

Where N1 is the number of primary turns

V1 is the primary voltage

N2 is the number of secondary turns

V2 is the secondary voltage

Step-up transformer:

A transformer that increases the voltage from primary to secondary (more secondary winding turns than primary winding turns) is called a step-up transformer.

Therefore, K > 1

Step down transformer:

A transformer that decreases the voltage from primary to secondary (more primary winding turns than secondary winding turns) is called a step-down transformer.

Therefore, K < 1

**Explanation:**

Transformation ratio (k) = \(\frac{N_2}{{{N_1}}} = \frac{1}{{{n}}} \)

When Port 2 inductance L is referred to Port 1

Then, inductance will be \(= \frac{L}{{{k^2}}} = \frac{L}{{\frac{1}{{{n^2}}}}} = {n^2}L\)

Identify the given circuit.

Option 2 : Step down transformer

Given circuit is the circuit of Half wave rectifier.

N_{1} > N_{2}

N_{1} = no. of turns in primary coil

N_{2} = no. of turns in secondary coil.

Let supply voltage V is sinusoidal

When 0 < α < π → V = +ve → so diode will conduct in forward Bias. Diode will be short circuited.

V_{0} = V_{2}

∵ \(\frac{{{V_1}}}{{{V_2}}} = \frac{{{N_1}}}{{{N_2}}}\)

\({V_2} = \frac{{{N_2}}}{{{N_1}}} \cdot {V_1}\)

**Case 2)** When π < α < 2π

Supply voltage will be negative so V_{2} will also be negative diode will conduct in Reverse bias. Diode will work as open circuit.

V_{0 }= 0

So output voltage wavefrom

It works like halw wave Rectrifier.

Transformer is a device which reduces or increases the AC voltage. The step-down transformer reduces the AC voltage from high to low whereas the step-up transformer increases the AC voltage from low to high. In half wave rectifier, we generally use a step-down transformer because the voltage needed for the diode is very small. Applying a large AC voltage without using transformer will permanently destroy the diode. So we use step-down transformer in half wave rectifier. In the step-down transformer, the primary winding has more turns than the secondary winding. So the step-down transformer reduces the voltage from primary winding to secondary winding.

The following figure shows the connection of an ideal transformer with primary to secondary turns ratio of 1:100. The applied primary voltage is 100 V (rms), 50 Hz, AC. The rms value of the current , in ampere, is __________.

Here \({{\rm{Z}}_{\rm{L}}}{\rm{\;}} = {\rm{\;R}} - {\rm{j}}{{\rm{X}}_{\rm{C}}}{\rm{\;}} = {\rm{\;}}\left( {80{\rm{\;}}-{\rm{\;j}}40} \right){\rm{\;k\Omega \;}}\)

Given \({{\rm{N}}_1}{\rm{\;}}:{\rm{\;}}{{\rm{N}}_2}{\rm{\;}} = {\rm{\;}}1{\rm{\;}}:{\rm{\;}}100\)

We know \(\frac{{{{\rm{Z}}_{{\rm{in}}}}}}{{{{\rm{Z}}_{\rm{L}}}}} = {\left( {\frac{{{{\rm{N}}_1}}}{{{{\rm{N}}_2}}}} \right)^2}\)

\(\therefore {{\rm{Z}}_{{\rm{in}}}} = \left( {80 - {\rm{j}}40} \right) \times {10^3} \times \frac{1}{{{{10}^4}}} = \left( {8 - {\rm{j}}4} \right){\rm{\Omega }}\)

Now the circuit becomes,

\(\therefore {\rm{I}} = \frac{{\rm{V}}}{{\rm{Z}}} = \frac{{100}}{{\left| {8 - {\rm{j}}4 + {\rm{j}}10} \right|}} = \frac{{100}}{{\left| {8 + {\rm{j}}6} \right|}} = \frac{{100}}{{10}} = 10\)

\(\therefore {\rm{\;}}{{\rm{I}}_{{\rm{RMS}}}}{\rm{\;}} = {\rm{\;}}10{\rm{\;A}}\)

For an ideal transformer, if I_{2 }/ I_{1} = 2 and E_{2} = 100 V, E_{1} will be:

Option 1 : 200 V

Concept:

Turn ratio (n) of a transformer is given as

\(n = \frac{{{N_1}}}{{{N_2}}} = \frac{{{E_1}}}{{{E_2}}} = \frac{{{I_2}}}{{{I_1}}}\)

Where,

N1 = Number of turns across primary side

N2 = Number of turns across secondary side

E1 = Primary side voltage

E2 = Secondary side voltage

I1 = Primary current

I2 = Secondary current

Calculation:

Given-

n = 2, E2 = 100 V

∴ \(2 = \frac{{{E_1}}}{{{100}}} \)

**E _{1} = 200 V**

Option 4 : 20(t - 1)V

__Concept:__

Farday’s laws of Electromagnetic Induction;

Statement1: There should be a change in flux linkage with the coil in order to induce emf in it.

Statement2: The magnitude of the induced emf is proportional to the rate of change of flux linkage.

Mathematically given by,

\(e = - N\frac{{d{\rm{\Psi }}}}{{dt}}\)

But \({\rm{\Psi }} = N\phi \) ⇒ flux linkages (Wb – turns)

∴ \(e = - N\frac{{d\phi }}{{dt}}\)

Where,

\({\rm{ }}\frac{{d\phi }}{{dt}}\) = Rate of changes in flux (Wb/sec)

N = Number of turns

e = induced emf in coil (volt)

__Calculation:__

Given,

N_{1} = 100 turns, N2 = 200 turns,

ϕm(t) = -0.05 (t2 – 2t)

Emf generated on secondary side

\(E =- N_2\frac{{d\phi }}{{dt}}\)

\(= - 200\frac{d}{{dt}}\left[ { - 0.05\left( {{t^2} - 2t} \right)} \right]\)

\( = 10\frac{d}{{dt}}\left( {{t^2} - 2t} \right)\)

= 10(2t - 1)

= 20(t -1)Consider the following statements regarding the ideal transformer:

1. The excitation current required to produce flux in the core is very less.

2. Iron losses in the core are zero.

3. The percentage voltage regulation of the ideal transformer is always zero.

4. Coupling coefficient K is unity between primary and secondary windings.

Which of the above statements are correct?

Option 4 : 2, 3 and 4 only

**Properties of Ideal Transformer:**

- The
**permeability**of the transformer core is**infinity**. - So, the
**excitation current**which is required to produce flux in the transformer core is**Zero.** - The
**magnetization curve is linear.** - So, the iron losses (both eddy current and hysteresis loss) in transformer core are zero.
**Resistance**of transformer**windings**is**zero.**- The entire flux of primary will links with the secondary winding, so,
**coupling coefficient K is unity.** - Hence,
**Magnetic leakage flux in transformer is completely zero.** - As magnetic leakage flux is zero, the
**leakage reactance of the transformer is zero.** - The resistance and reactance both are zero means the impedance will be zero.
- As leakage impedance is zero, there is
**no leakage impedance drop in the transformer**and hence the**percentage voltage regulation of the transformer is zero.**

What is the power dissipated in the primary of the ideal transformer in the given circuit ?

Option 4 : 62.5 W

__Concept:__

Power in an electric circuit is a combination of the Voltage and Current in a circuit.

Power in an electrical circuit is defined as:

P = V × I

\(P = {I^2}R = \frac{{{V^2}}}{R}\)

Turns ratio in the transformer

It is defined as the number of turns on its secondary divided by the number of turns in the primary coil.

\(\frac{{{N_2}}}{{{N_1}}} = \frac{{{V_2}}}{{{V_1}}} = \frac{{{I_1}}}{{{I_2}}}\)

N2: Turns in the secondary coil

N1: Turns in the primary coil

V2: Secondary voltage

V1: Primary voltage

I1: Primary current

I2: Secondary current

__Calculation:__

Consider the circuit as shown:

Based on the turns ratio formula the voltage at the secondary winding is:

V2 = 5 × 50

V2 = 250 V

Secondary current is:

\({I_2} = \frac{{250}}{{1000}} = 0.25A\)

Primary current is:

\({I_1} = {I_2}\frac{{{N_2}}}{{{N_1}}}\)

\({I_1} = 0.25 \times \frac{5}{1}\)

I1 = 1.25 A

Power dissipated in the primary is:

P = 50 × 1.25

P = 62.5 W

**Alternative method:**

Referring secondary load to the primary

\(R_2' = \frac{R_2}{K^2}\)

K: Turns ratio

\(R_2' = \frac{1k\Omega}{25} = 40 \Omega\)

I1 = 50∠0° / 40

I1 = 1.25 A

P = I12 × R

P = 1.252 × 40

P = 62.5 W

Option 1 : coper loss = iron loss

Copper loss:

A loss in a transformer that takes place in the winding resistance of a transformer is known as copper loss. These are variable losses.

Copper loss = i2R

Iron loss:

Iron losses = Eddy current loss + hysteresis loss

Eddy current loss = Ket2f2B2

Hysteresis loss = KhfBx

These are constant losses.

Efficiency:

The efficiency of the Transformer \((\eta ) = \frac{{XS\cos \phi }}{{XS\cos \phi + {P_i} + {X^2}{P_e}}}\)

Where, X = Fraction of load

S = Apparent power in kVA

Wi = Iron losses

Wcu = Copper losses

The transformer will give the maximum efficiency when their copper loss is equal to the iron loss.

Efficiency is maximum at some fraction x of full load.

\(x = \sqrt {\frac{{{W_i}}}{{{W_{cu}}}}}\)

Where Wi = iron losses

Wcu = copper losses

kVA at maximum efficiency is given by,

\(kVA\;at\;{\eta _{max}} = full\;load\;kVA \times \sqrt {\frac{{{W_i}}}{{{W_{cu}}}}}\)