EM-Problems Second Semester 2018-2019

1

EM-Problems Second Semester 2018-2019

Problem-1

A transformer has 500 primary turns and 3000 secondary turns. If the primary voltage is 240 V,

determine the Secondary voltage, assuming an ideal transformer.

Solution:

For an ideal transformer, voltage ratio = turns ratio, i.e.

𝑉1

𝑉2=

𝑁1

𝑁2, ℎ𝑒𝑛𝑐𝑒

240

𝑉2=

500

3000

𝑇ℎ𝑢𝑠 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑉2 = (3000) (240)

(500)= 1440 𝑉 𝑜𝑟 1.44 𝑘𝑉

Problem-2

An ideal transformer with a turns ratio of 2:7 is fed from a 240 V supply. Determine its output

voltage.

Solution:

A turns ratio of 2:7 means that the transformer has 2 turns on the primary for every 7 turns on the

secondary (i.e. a step – up transformer). Thus,

𝑁1

𝑁2=

2

7

𝐹𝑜𝑟 𝑎𝑛 𝑖𝑑𝑒𝑎𝑙 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟,𝑁1

𝑁2=

𝑉1

𝑉2 ; ℎ𝑒𝑛𝑐𝑒

2

7=

240

𝑉2

𝑇ℎ𝑢𝑠 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑉2 = (240) (7)

(2)= 840 𝑉

Problem-3

An ideal transformer has a turns ratio of 8:1 and the Primary current 3 A when it is supplied at 240

V. Calculate the secondary voltage and current.

Solution:

𝑨 𝒕𝒖𝒓𝒏𝒔 𝒓𝒂𝒕𝒊𝒐 𝒐𝒇 𝟖: 𝟏 𝒎𝒆𝒂𝒏𝒔 𝑵𝟏

𝑵𝟐=

𝟖

𝟏, 𝒊. 𝒆. 𝒂 𝒔𝒕𝒆𝒑 − 𝒅𝒐𝒘𝒏 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒆𝒓.

𝑵𝟏

𝑵𝟐=

𝑽𝟏

𝑽𝟐 𝒐𝒓 𝒔𝒆𝒄𝒐𝒏𝒅𝒂𝒓𝒚 𝒗𝒐𝒍𝒕𝒂𝒈𝒆 𝑽𝟐 = 𝑽𝟏 (

𝑵𝟐

𝑵𝟏) = 𝟐𝟒𝟎 (

𝟏

𝟖) = 𝟑𝟎 𝑽𝒐𝒍𝒕𝒔

2

Problem-4

An ideal transformer, connected to 240 V mains, supplies a 12 V, 150 W lamp. Calculate the

transformer turns ratio and the current taken from the supply.

Solution:

𝑉1 = 240𝑉, 𝑉2 = 12𝑉, 𝐼2 = 𝑃

𝑉2=

150

12= 12.5 𝐴

𝑇𝑢𝑟𝑛𝑠 𝑟𝑎𝑡𝑖𝑜 = 𝑁1

𝑁2=

𝑉1

𝑉2=

240

12= 20

𝑉1

𝑉2=

𝐼2

𝐼1, 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝐼1 = 𝐼2 (

𝑉2

𝑉1) = 12.5 (

12

240)

𝐻𝑒𝑛𝑐𝑒 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑡𝑎𝑘𝑒𝑛 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑠𝑢𝑝𝑝𝑙𝑦, 𝐼1 = 12.5

20= 0.625 𝐴

Problem-5

A 5-kVA single-phase transformer has a turns ratio of 10:1 and is fed from a 2.5 kV supply.

Neglecting losses, determine:

a) the full-load secondary current.

b) the minimum load resistance which can be connected a cross the secondary winding to give

full load kVA.

c) the primary current at full load kVA.

Solution: 𝑁1

𝑁2=

10

1 𝑎𝑛𝑑 𝑉1 = 2.5 𝑘𝑉 = 2500 𝑉

𝑆𝑖𝑛𝑐𝑒 𝑁1

𝑁2=

𝑉1

𝑉2, 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑣𝑜𝑙𝑡𝑎𝑔𝑒

𝑉2 = 𝑉1 (𝑁2

𝑁1) = 2500 (

1

10) = 250 𝑉

𝑇ℎ𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟 𝑟𝑎𝑡𝑖𝑛𝑔 𝑖𝑛 𝑣𝑜𝑙𝑡 − 𝑎𝑚𝑝𝑒𝑟𝑒𝑠 = 𝑉2 𝐼2 (𝑎𝑡 𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑), 𝑖. 𝑒. 5000 = 250 𝐼2

𝐻𝑒𝑛𝑐𝑒 𝑓𝑢𝑙𝑙 𝑙𝑜𝑎𝑑 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝐼2 = 5000

250= 20 𝐴

𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑙𝑜𝑎𝑑 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒, 𝑅𝐿 = 𝑉2

𝐼2=

250

20= 12.5Ω

𝑁1

𝑁2=

𝐼2

𝐼1, 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑐𝑢𝑟𝑟𝑒𝑛𝑡, 𝐼1 = 𝐼2 (

𝑁2

𝑁1) = 20 (

1

10) = 2 𝐴

3

Problem-6

A 2400 V/400 V single-phase transformer takes a no-load current of 0.5 A and the core

losses 400 W.

Determine the values of the magnetizing and core loss components of the no-load current

(Io or Ie). Draw to scale the no-load phasor diagram for the transformer.

Solution:

𝑉1 = 2400 𝑉, 𝑉2 = 400 𝑉, 𝐼𝑜 = 0.5 𝐴

The no-load phasor diagram is shown in fig. below

4

Problem-7

A transformer takes a current of 0.8 A when its primary is connected to a 240 volt, 50 Hz supply, the

secondary being on open circuit. If the power absorbed is 72 watts, determine

a) The iron loss current.

b) The power factor on no-load.

c) The magnetizing current.

Solution:

Io = 0.8 A, V = 240 V

a) Power absorbed = total core loss = 72 Watt = V1 Io cos ϕo

Hence 72 = 240 Io cos ϕo

And iron loss current, 𝐼𝑐 = 𝐼𝑜 cos Φ𝑜 = 72

240= 0.3 𝐴

𝑃𝑜𝑤𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟 𝑎𝑡 𝑛𝑜 𝑙𝑜𝑎𝑑, cos Φ𝑜 = 𝐼𝑐

𝐼𝑜=

0.3

0.8= 0.375

𝐼𝑜2 = 𝐼𝑐

2 + 𝐼𝑀2

𝐼𝑀 = √(𝐼𝑜2 − 𝐼𝑐

2)

𝐼𝑀 = √(0.82 − 0.32)

𝐼𝑀 = 0.74 𝐴

Фm

IC

IM

IO = Ie

E2

E1

5

Problem-8 A 100 kVA, 4000 V/200 V, 50 Hz single-phase transformer has100 secondary turns. Determine

a) The primary and secondary current.

b) The number of primary turns.

c) The maximum value of the flux.

Note:

The magnetic flux set up in the core of a transformer when an alternating voltage is applied to its

primary winding is also alternating and is sinusoidal.

Let ɸm be the maximum value of the flux

and f be the frequency of the supply. The

time for 1 cycle of the alternating flux is

the periodic time T, where T = 1/f

seconds

The flux rises sinusoidally from zero to

its maximum value in ¼ cycle, and the

time for ¼ cycle is ¼ f seconds.

Hence the average rate of change of flux =

ɸm / (1/4 f) = 4 f ɸm Wb/S and since 1Wb/s

= 1volt, the average e.m.f. induced in each turn = 4 f ɸm volts.

As the flux varies sinusoidally, then a sinusoidal e.m.f. will be induced in each turn of both primary

and secondary windings.

For a sine wave, form factor = rms value/average value = 1.11

Hence rms value = form factor * average value = 1.11 * average value

Thus rms e.m.f. induced in each turn = 1.11 * 4 f ɸm volts = 4.44 f ɸm volts

Therefore, rms value of e.m.f. induced in primary,

E1 = 4.44 f ɸm N1 volts

And rms value of e.m.f. induced in secondary,

E2 = 4.44 f ɸm N2 volts

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Solution:

V1 = 400V, V2 = 200 V, f = 50 Hz, N2 = 100 turns

a) Transformer rating = V1 * I1 = V2 *I2 = 100000 VA

hence primary current, 𝐼1 = 100000

𝑉1=

100000

4000= 25 𝐴

and secondary Current, 𝐼2 = 100000

𝑉2=

100000

200= 500 𝐴

from which, primary turns, 𝑁1 = (𝑉1

𝑉2) (𝑁2) = (

4000

200) (100)

b) i.e., N1 = 2000 turns

c)

𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝑚𝑎𝑥𝑚𝑖𝑚𝑢𝑚 𝑓𝑙𝑢𝑥 Φ𝑚 = 𝐸2

4.44 𝑓 𝑁2=

200

4.44 (50) (100)

(assuming E2 = V2)

𝜙𝑚 = 9.01 ∗ 10−3 𝑊𝑏 𝑜𝑟 9.01 𝑚𝑊𝑏

Where E1 = 4.44 f Φm N1

From which, 𝜙𝑚 = 𝐸1

4.44 𝑓 𝑁1=

4000

4.44 (50) (2000)

(assuming E1 = V1)

𝜙𝑚 = 9.01 ∗ 10−3 𝑊𝑏 𝑜𝑟 9.01 𝑚𝑊𝑏

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Problem-9

A single-phase, 50 Hz transformer has 25 primary turns and 300 secondary turns. The cross-

sectional area of the core is 300 cm2. When the primary winding is connected to a 250 V supply,

determine

a) The maximum value of the flux density in the core.

b) The voltage induced in the secondary winding.

Solution:

e.m.f. E1 = 4.44 f ɸm N1 volts i.e.,

250 = 4.44 (50) Φm (25)

𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑓𝑙𝑢𝑥 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 𝜙𝑚 = 250

(4.44) (50) (25) 𝑊𝑏 = 0.04505 𝑊𝑏

However,𝜙𝑚 = 𝐵𝑚 ∗ 𝐴, 𝑤ℎ𝑒𝑟𝑒 𝐵𝑚 = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑓𝑙𝑢𝑥 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑜𝑟𝑒 𝑎𝑛𝑑

𝐴 = 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑟𝑒.

Hence Bm * 300 * 10-4

= 0.04505

From which, maximum flus density, 𝐵𝑚 = 0.04504

300∗ 10−4 = 1.5 𝑇

b) 𝑉1

𝑉2=

𝑁1

𝑁2, 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝑉2 = 𝑉1 (

𝑁2

𝑁1)

𝑖. 𝑒. , 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑤𝑖𝑛𝑑𝑖𝑛𝑔,

𝑉2 = (250) (300

25) = 3000 𝑉 𝑜𝑟 3 𝑘𝑉

8

Problem-10

A single-phase 500 V/100 V, 50 Hz transformer has a maximum core flux density of 1.5 T and an

effective core cross-sectional area of 50 cm2. Determine the number of primary and secondary turns.

Solution:

The e.m.f. Equation for a transformer is E = 4.44 f ɸm N

And maximum flux, 𝜙𝑚 = 𝐵 ∗ 𝐴 = (1.5) (50 ∗ 10−4) = 75 ∗ 10−4 𝑊𝑏

Since E1 = 4.44 f ɸm N1

Then primary turns,𝑁1 = 𝐸1

4.44 𝑓 ∅𝑚=

500

4.44 (50) (75∗ 10−4)= 300 𝑡𝑢𝑟𝑚𝑠

Since E2 = 4.44 f ɸm N2

Then secondary turns, 𝑁2 = 𝐸2

4.44 𝑓 ∅𝑚=

500

4.44 (50) (75∗ 10−4)= 60 𝑡𝑢𝑟𝑚𝑠

Problem-11

A 4500 V/225 V, 50 Hz single-phase transformer is to have an approximate e.m.f. per turn of 15 V

and operate with a maximum flux of 1.4 T. Calculate

a) The number of primary and secondary turns.

b) The cross-sectional area of the core.

Solution:

a) 𝐸. 𝑚. 𝑓. 𝑝𝑒𝑟 𝑡𝑢𝑟𝑛 = 𝐸1

𝑁1=

𝐸2

𝑁2= 15

𝐻𝑒𝑛𝑐𝑒 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑡𝑢𝑟𝑛𝑠, 𝑁1 = 𝐸1

15=

4500

15= 300

𝑎𝑛𝑑 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑡𝑢𝑟𝑛𝑠, 𝑁2 = 𝐸2

15=

225

15= 15

b) E.m.f. E1 = 4.44 f ɸm N1

from which,Φ𝑚 = 𝐸1

4.44 𝑓 𝑁1=

4500

4.44 (50) (300)= 0.0676 𝑊𝑏

𝑁𝑜𝑤 𝑓𝑙𝑢𝑥 𝜙𝑚 = 𝐵𝑚 ∗ 𝐴, 𝑤ℎ𝑒𝑟𝑒 𝐴 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 − 𝑆𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑟𝑒, ℎ𝑒𝑛𝑐𝑒

𝑎𝑟𝑒𝑎 𝐴 = 𝜙𝑚

𝐵𝑚=

0.0676

1.4= 0.0483 𝑚2 𝑜𝑟 483 𝑐𝑚2

9

Problem-12

A single-phase transformer has 2000 turns on the primary and 800 turns on the secondary. Its no

load current is 5 A at a power factor of 0.20 lagging. Assuming the volt drop in the windings is

negligible, determine the primary current and power factor when the secondary current is 100 A at a

power factor of 0.85 lagging.

Solution:

Let 𝑰𝟏′ be the component of the primary current which provides the

restoring mmf. Then

𝐼1′ 𝑁1 = 𝐼2 𝑁2

i.e.,

𝐼1′ (2000) = (100) (800)

𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝐼1′ =

(100) (800)

2000= 40 𝐴

If the power factor of the secondary is 0.85 the cos ϕ2 = 0.85, from which, ϕ2 = arccos 0.85 = 31.8o.

If the power factor on no load is 0.2, then cos ϕ0 = 0.2, and ϕ0 = arccos 0.2 = 78.5o. In the phasor

diagram shown in fig. above I2 = 100 A is shown at an angle of ϕ2 = 31.8o

to V2 and 𝑰𝟏′ = 40 A is

shown in anti-phase to I2

The no load current I0 = 5 A is shown at an angle of ϕ0 = 78.5o to V1.

Current I1 is the phasor sum of 𝑰𝟏′ and I0 and by drawing to scale, I1 = 44 A and angle ϕ1= 37

o

𝐵𝑦 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛, 𝐼1 cos 𝜙1 = 𝑜𝑎 + 𝑜𝑏 = 𝐼𝑜 cos 𝜙𝑜 + 𝐼1′ cos 𝜙2

= (5) ∗ (0.2) + (40) ∗ (0.85) = 𝟑𝟓. 𝟎 𝑨

𝑎𝑛𝑑 𝐼1 sin 𝜙1 = 𝑜𝑐 + 𝑜𝑑 = 𝐼𝑜 sin 𝜙𝑜 + 𝐼1′ sin 𝜙2

= (5) sin 78.5𝑜 + (40) sin 31.8 𝑜 = 𝟐𝟓. 𝟗𝟖 𝑨

𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝐼1 = √(35.02 + 25.982) = 𝟒𝟑. 𝟓𝟗 𝑨

𝑎𝑛𝑑 tan 𝜙1 = (25.98

35.0) , 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ,

𝜙1 = arctan (25.98

35.0) = 𝟑𝟔. 𝟓𝟗𝒐

𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑝𝑜𝑤𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 = cos 𝜙1 = cos 36.59𝑜 = 𝟎. 𝟖 𝟎

10

Problem-13

A transformer has 600 primary turns and 150 secondary turns. The primary and secondary

resistances are 0.25 Ω and 0.01 Ω respectively and the corresponding leakage reactances are 1.0 Ω

and 0.04 Ω respectively. Determine:

a) The equivalent resistance referred to the primary winding

b) The equivalent reactance referred to the primary winding

c) The equivalent impedance referred to the primary winding

d) The phase angle of the impedance.

Solution:

𝑅𝑒𝑞 = 𝑅1 + 𝑅2 (𝑉1

𝑉1)

2

𝑅𝑒𝑞 = 0.25 + 0.01 (600

150)

2

= 0.41 Ω

𝑋𝑒𝑞 = 𝑋1 + 𝑋2 (𝑉1

𝑉1)

2

𝑅𝑒𝑞 = 1.0 + 0.04 (600

150)

2

= 1.64 Ω

𝑍𝑒 = √(𝑅𝑒2 + 𝑋𝑒

2) = 1.69 Ω

𝜙𝑒 = 𝐴𝑟𝑐𝑐𝑜𝑠 (0.41

1.69) = 75.96𝑜

Problem-14

A 5 kVA, 200V/400V, single-phase transformer has a secondary terminal voltage of 387.6 volts

when loaded. Determine the regulation of the transformer.

Solution:

𝑉𝑅 = 𝑁𝑜 𝑙𝑜𝑎𝑑 𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 − 𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑜𝑛 𝑙𝑜𝑎𝑑

𝑛𝑜 𝑙𝑜𝑎𝑑 𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 ∗ 100%

𝑉𝑅 = 400 − 387.6

400 ∗ 100% = 3.1 %

11

Problem-15

The equivalent circuit impedances of a 20-kVA, 8000/240 V, 60-Hz transformer are to be

determined. The open-circuit test was performed on the secondary side of the transformer (to reduce

the maximum voltage to be measured) and the short circuit test were performed on the primary side

of the transformer (to reduce the maximum current to be measured). The following data were taken:

Open – Circuit test

(on secondary)

Short – Circuit test

(on Primary)

VOC = 240 V VSC = 489 V

IOC = 7.133 A ISC = 2.5 A

POC = 400 W PSC = 240 W

Find the impedances of the approximate equivalent circuit referred to the primary side, and

sketch that circuit.

Solution:

The turns ratio of this transformer is a = 8000/240 = 33.3333. The power factor during the open-

circuit test is

Therefore, the values of the excitation branch referred to the low-voltage (secondary) side are

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