Two coils, one iron core, no moving parts. This file shows how a transformer trades current for voltage without creating power, why it needs AC, where it loses energy, and why power lines run at such high voltage.
Transformer. Nothing moves and no wire joins the two coils. The alternating current in the primary makes an alternating flux in the iron core, and that flux induces an emf in every turn of the secondary.
Picture it
A bicycle's gears let you trade speed for effort: a low gear makes pedalling easy but slow, a high gear makes it hard but fast. You cannot get more energy out than your legs put in. A transformer is a gearbox for electricity. It trades current for voltage, but never gives out more power than it takes in.
In exam language
A transformer works on mutual induction. It changes the voltage of an alternating supply and transfers electrical power from one circuit to another without changing the frequency. It has two parts: the windings (primary and secondary copper coils, electrically insulated from each other) and a soft iron core that links them magnetically.
Type
Turns
Voltage
Current
Step-up
Nₛ > Nₚ
Raised
Lowered
Step-down
Nₛ < Nₚ
Lowered
Raised
The ideal transformer relations
Turns ratio. In an ideal transformer, raising the turns ratio raises the voltage and lowers the current by the same factor; the power out stays equal to the power in.
An ideal transformer has three assumptions, and each gives one relation:
Assumption
Consequence
Relation
No flux leakage
Every turn of both coils carries the same flux, so the same emf per turn
Eₛ/Eₚ = Nₛ/Nₚ
No load resistance losses (coils of negligible resistance)
Terminal voltages equal the emfs
Vₛ/Vₚ = Nₛ/Nₚ
No power loss
Pₒᵤₜ = Pᵢₙ, so VₛIₛ = VₚIₚ
Iₛ/Iₚ = Nₚ/Nₛ
Vₛ/Vₚ = Nₛ/Nₚ = Iₚ/Iₛ
Trap
Current goes the opposite way to voltage: the coil with more turns has the higher voltage and the smaller current. Writing Iₛ/Iₚ = Nₛ/Nₚ is a ratio reversal, one of her recurring error types.
Unless a question says otherwise, treat a transformer as ideal (Pₒᵤₜ = Pᵢₙ).
What a transformer cannot do
Statement
Reason
It cannot work on a DC supply
A steady current gives a steady flux, so no emf is induced in the secondary (except a brief kick at switching)
It is not an amplifier
It cannot increase power; a transistor can have a power gain, a transformer cannot
It does not change frequency
The secondary emf follows the same alternating flux
It has no mechanical losses
It has no moving parts, which is why its efficiency is higher than that of motors and generators
Real transformers: efficiency and losses
efficiency η = Pₒᵤₜ/Pᵢₙ × 100% = VₛIₛ/VₚIₚ × 100%
Loss
Where and why
Remedy
Copper (Joule heating)
Both coils; I²R heating
Thick wire for the high-current coil; circulating oil for cooling
Flux leakage
Between the coils; some flux misses the secondary
Wind both coils tightly on a common soft iron core (coupling factor close to 1)
Hysteresis (iron loss)
The core is magnetised and demagnetised every cycle
The high-voltage coil has many turns of thin wire; the high-current coil has few turns of thick wire.
Why power lines use very high voltage
Same power, different voltage. Power = VI, so high voltage means small current, and the heat wasted in the line, I²R, falls with the square of the current.
line current I = P/V voltage drop = IR line loss = I²R = (P/V)² R
So power stations use step-up transformers to send power at very high voltage and low current, and step-down transformers near homes bring the voltage back down.
Allen examples, checked
Illustration
Working
Check
38: 220 V → 2200 V, Nₛ = 2000
Nₚ = Nₛ × Vₚ/Vₛ = 2000 × 220/2200 = 200
Correct (200)
39: 2300 V → 230 V, Nₚ = 4000
Nₛ = 4000 × 230/2300 = 400
Correct
40: 240 V → 24 V, lights a 24 V, 24 W bulb
Iₚ = Pₛ/Vₚ = 24/240 = 0.1 A
Correct
41: 100 and 300 turns, input 60 W
Ideal: output 60 W
Correct
42: 100 W, 110 V bulb from 220 V, 0.5 A
η = 100/(220 × 0.5) = 90.9%
Answer correct, but the solution first writes Pᵢₙ = Pₒᵤₜ, which contradicts η < 100%
43: 50 and 1500 turns, flux φ = (2 + 4t) Wb
If φ is the primary's total flux linkage: Eₚ = 4 V, Vₛ = 4 × 1500/50 = 120 V
Allen's 120 V needs the linkage reading; if φ is flux per turn, Eₚ = 50 × 4 = 200 V and Vₛ = 6000 V
44: 2500 V input, 20 : 1, η = 100%
Vₛ = 2500/20 = 125 V (the 80 A output current is not needed)
Correct
Correction to the Allen module
Illustration 42 states Pin = Pout and then computes an efficiency of 90.9%. Both cannot be true: the input is 220 × 0.5 = 110 W and the output 100 W, so 10 W is lost. Illustration 43 does not say whether φ is the flux per turn or the total flux linkage of the primary; the printed answer uses total linkage. In an exam, check which reading the options support.
Formula sheet
Formula
Meaning
When to use
Watch out
Vₛ/Vₚ = Nₛ/Nₚ
Voltage ratio
Any (ideal) transformer
Step-up if Nₛ > Nₚ
Iₛ/Iₚ = Nₚ/Nₛ
Current ratio
Ideal transformer
Inverse of the turns ratio
VₛIₛ = VₚIₚ
Power balance
Ideal transformer
Only if η = 100%
η = VₛIₛ/VₚIₚ
Efficiency
Real transformer
× 100 for a percentage
Eₚ = Nₚ dφ/dt
Primary emf from per-turn flux
Flux per turn given
If total linkage is given, no N
I = P/V, loss = I²R
Transmission line
Power sent over a line
Loss ∝ 1/V² at fixed P
Pₗₒₛₛ = Pᵢₙ − Pₒᵤₜ
Heat lost
Given efficiency
Loss = (1 − η)Pᵢₙ
Allen pages 134–139 checked
Item
Check
Principle (mutual induction), parts, AC only, not an amplifier
Correct
Step-up and step-down; transmission at high voltage
Correct
Ideal relations Vₛ/Vₚ = Nₛ/Nₚ = Iₚ/Iₛ
Correct
Losses and remedies (copper, flux leakage, hysteresis, eddy)
Correct
Illustrations 38, 39, 40, 41, 44
Correct
Illustration 42
Answer right; Pin = Pout line contradicts it
Illustration 43
Answer needs φ read as total flux linkage
Beginner's Box 9 answer key
Q
Answer
1
To reduce eddy current losses in the core. Thin insulated sheets confine the eddy currents to small, high-resistance paths.
(Reading the question as 'Why can't a transformer step up a DC voltage?') A steady current gives a steady flux in the core, so no emf is induced in the secondary.
4
Any two of: induction furnace, electromagnetic brakes, dead-beat galvanometer (electromagnetic damping), speedometer, induction motor, diathermy.
NEET practice: 35 questions
Ratio reversal is the main trap here: voltage follows the turns, current goes against them. Every number was recalculated in Python.
Q1Numerical
An ideal transformer has 100 primary turns and 500 secondary turns. The primary is connected to 220 V AC. The secondary voltage is
(A)5500 V
(B)44 V
(C)220 V
(D)1100 V
Show the solution
Given
Nₚ = 100, Nₛ = 500, Vₚ = 220 V
Asked
Vₛ
Concept
Voltage ratio.
Formula
Vₛ = Vₚ Nₛ/Nₚ
Baby steps
Nₛ/Nₚ = 5.
Vₛ = 1100 V.
Answer
(D) 1100 V
Why not the others
44 V inverts the ratio. 220 V ignores it. 5500 V squares it.
Shortcut
More turns, more volts.
Where it went wrong
Ratio reversal.
Q2Numerical
A step-down transformer changes 230 V to 11.5 V. Its primary has 2000 turns. The number of secondary turns is
(A)200
(B)40 000
(C)100
(D)20
Show the solution
Given
Vₚ = 230 V, Vₛ = 11.5 V, Nₚ = 2000
Asked
Nₛ
Concept
Voltage ratio.
Formula
Nₛ = Nₚ Vₛ/Vₚ
Baby steps
Vₛ/Vₚ = 1/20.
Nₛ = 2000/20 = 100.
Answer
(C) 100
Why not the others
40 000 inverts. 200 and 20 are a factor of 2 or 5 out.
Shortcut
Step-down → fewer secondary turns.
Where it went wrong
Multiplying by 20.
Q3Numerical
An ideal transformer steps 220 V down to 22 V. The secondary current is 5 A. The primary current is
(A)5 A
(B)50 A
(C)0.5 A
(D)0.05 A
Show the solution
Given
Vₚ = 220 V, Vₛ = 22 V, Iₛ = 5 A
Asked
Iₚ
Concept
Power balance.
Formula
VₚIₚ = VₛIₛ
Baby steps
VₛIₛ = 110 W.
Iₚ = 110/220 = 0.5 A.
Answer
(C) 0.5 A
Why not the others
50 A applies the voltage ratio the wrong way. 5 A ignores the ratio. 0.05 A divides by 100.
Shortcut
Current goes the opposite way to voltage.
Where it went wrong
Ratio reversal.
Q4Numerical
An ideal transformer connected to 240 V mains lights a 12 V, 60 W lamp at full brightness. The primary current is
(A)0.25 A
(B)5 A
(C)20 A
(D)0.05 A
Show the solution
Given
Vₚ = 240 V, lamp 60 W
Asked
Iₚ
Concept
Ideal: input power = output power.
Formula
Iₚ = Pₛ/Vₚ
Baby steps
Pₛ = 60 W.
Iₚ = 60/240 = 0.25 A.
Answer
(A) 0.25 A
Why not the others
5 A is the lamp current (60/12). 20 A is 240/12. 0.05 A divides by 1200.
Shortcut
Power over primary voltage.
Where it went wrong
Giving the secondary current.
Q5Numerical
A transformer draws 2 A from a 220 V supply and delivers 9 A at 44 V. Its efficiency is
(A)90%
(B)80%
(C)100%
(D)95%
Show the solution
Given
Vₚ = 220 V, Iₚ = 2 A, Vₛ = 44 V, Iₛ = 9 A
Asked
η
Concept
Efficiency = output power/input power.
Formula
η = VₛIₛ/VₚIₚ
Baby steps
Pₒᵤₜ = 396 W.
Pᵢₙ = 440 W.
η = 0.9 = 90%.
Answer
(A) 90%
Why not the others
80% and 95% miscalculate. 100% assumes an ideal transformer.
Shortcut
Output ÷ input.
Where it went wrong
Assuming ideal when data show a loss.
Q6Numerical
A step-down transformer reduces 220 V to 22 V. The primary draws 1 A and the secondary delivers 8 A. The efficiency is
(A)125%
(B)90%
(C)100%
(D)80%
Show the solution
Given
Vₚ = 220 V, Iₚ = 1 A, Vₛ = 22 V, Iₛ = 8 A
Asked
η
Concept
Efficiency.
Formula
η = VₛIₛ/VₚIₚ
Baby steps
Pₒᵤₜ = 176 W.
Pᵢₙ = 220 W.
η = 80%.
Answer
(D) 80%
Why not the others
90% is Beginner's Box 9 Q2's answer. 100% assumes ideal. 125% is impossible.
Shortcut
Output ÷ input.
Where it went wrong
Copying a remembered answer.
Q7Numerical
A transformer of efficiency 80% takes 500 W from the mains. The power it delivers is
(A)400 W
(B)625 W
(C)100 W
(D)500 W
Show the solution
Given
η = 0.8, Pᵢₙ = 500 W
Asked
Pₒᵤₜ
Concept
Pₒᵤₜ = ηPᵢₙ.
Formula
Pₒᵤₜ = ηPᵢₙ
Baby steps
0.8 × 500 = 400 W.
Answer
(A) 400 W
Why not the others
625 W divides by η. 100 W is the loss. 500 W assumes ideal.
Shortcut
Output is less than input.
Where it went wrong
Dividing by the efficiency.
Q8Numerical
The total flux linkage of the primary coil (100 turns) of an ideal transformer is (3 + 2t) Wb. The secondary has 400 turns. The secondary voltage is
(A)0.5 V
(B)2 V
(C)800 V
(D)8 V
Show the solution
Given
Nₚφ = 3 + 2t (total linkage), Nₚ = 100, Nₛ = 400
Asked
Vₛ
Concept
Given total linkage, the emf needs no extra N.
Formula
Eₚ = d(Nₚφ)/dt, Vₛ = Eₚ Nₛ/Nₚ
Baby steps
Eₚ = 2 V.
Vₛ = 2 × 400/100 = 8 V.
Answer
(D) 8 V
Why not the others
2 V is the primary voltage. 800 V multiplies the linkage rate by Nₛ directly. 0.5 V inverts the ratio.
Shortcut
Linkage given → e = its rate.
Where it went wrong
Multiplying total linkage by N again.
Q9Numerical
In an ideal transformer the flux through each turn of the core windings is 0.02t Wb. The secondary has 500 turns. The secondary voltage is
(A)10 V
(B)0.02 V
(C)500 V
(D)25 000 V
Show the solution
Given
φ per turn = 0.02t, Nₛ = 500
Asked
Vₛ
Concept
Every turn of the secondary carries the same per-turn flux.
Formula
Eₛ = Nₛ dφ/dt
Baby steps
dφ/dt = 0.02 V per turn.
Vₛ = 500 × 0.02 = 10 V.
Answer
(A) 10 V
Why not the others
0.02 V is one turn. 500 V drops the 0.02. 25 000 V divides instead of multiplying.
Shortcut
Per-turn flux × turns.
Where it went wrong
Forgetting N when the flux is per turn.
Q10Concept
10 kW is sent through a line of total resistance 0.5 Ω, first at 250 V, then at 2500 V. The power lost in the line at 2500 V is what fraction of that at 250 V?
(A)1/100
(B)1/10
(C)10
(D)1
Show the solution
Given
Same P, R; V × 10
Asked
Loss ratio
Concept
Loss = (P/V)²R.
Formula
Loss ∝ 1/V²
Baby steps
At 250 V: I = 40 A, loss = 800 W.
At 2500 V: I = 4 A, loss = 8 W.
Ratio 1/100.
Answer
(A) 1/100
Why not the others
1/10 forgets to square. 10 inverts. 1 ignores the change.
Shortcut
Voltage × 10 → loss ÷ 100.
Where it went wrong
Not squaring the current.
Q11Numerical
A power station sends 1 MW at 20 kV. The current in the transmission line is
(A)50 000 A
(B)20 A
(C)50 A
(D)0.05 A
Show the solution
Given
P = 10⁶ W, V = 2 × 10⁴ V
Asked
I
Concept
I = P/V.
Formula
I = P/V
Baby steps
10⁶/(2 × 10⁴) = 50 A.
Answer
(C) 50 A
Why not the others
20 A uses the kV number. 50 000 A uses V = 20. 0.05 A uses P = 1 kW.
Shortcut
Watts over volts.
Where it went wrong
Unit prefixes.
Q12Concept
The same power is transmitted through the same line, but at twice the voltage. The power lost in the line becomes
(A)unchanged
(B)one half
(C)twice
(D)one quarter
Show the solution
Given
V → 2V, P and R fixed
Asked
New loss
Concept
Loss = (P/V)²R.
Formula
Loss ∝ 1/V²
Baby steps
Current halves.
I² falls to a quarter.
Answer
(D) one quarter
Why not the others
One half forgets to square. Twice inverts. Unchanged ignores the current.
Shortcut
Square of the current.
Where it went wrong
Using loss ∝ V²/R with the line voltage.
Q13Concept
In an ideal transformer the primary current is 10 A and the secondary current is 2 A. The ratio Nₛ : Nₚ is
(A)1 : 5
(B)5 : 1
(C)25 : 1
(D)1 : 1
Show the solution
Given
Iₚ = 10 A, Iₛ = 2 A
Asked
Nₛ : Nₚ
Concept
Current ratio is the inverse of the turns ratio.
Formula
Nₛ/Nₚ = Iₚ/Iₛ
Baby steps
Iₚ/Iₛ = 5.
Nₛ : Nₚ = 5 : 1 (step-up).
Answer
(B) 5 : 1
Why not the others
1 : 5 is the ratio reversal. 25 : 1 squares. 1 : 1 ignores the currents.
Shortcut
Smaller current → more turns.
Where it went wrong
Ratio reversal.
Q14Numerical
The input voltage of an ideal transformer is 1100 V and the turns ratio Nₚ : Nₛ is 10 : 1. The output current is 50 A. The output voltage is
(A)22 V
(B)11 000 V
(C)110 V
(D)5 V
Show the solution
Given
Vₚ = 1100 V, Nₚ : Nₛ = 10 : 1, Iₛ = 50 A
Asked
Vₛ
Concept
Voltage ratio; the output current is extra information (Allen Illustration 44 pattern).
Formula
Vₛ = Vₚ Nₛ/Nₚ
Baby steps
Vₛ = 1100/10 = 110 V.
Answer
(C) 110 V
Why not the others
11 000 V inverts. 22 V divides by 50. 5 V divides 50 by 10.
Shortcut
Ignore data you don't need.
Where it went wrong
Using the output current somewhere.
Q15Numerical
An ideal transformer with Nₚ : Nₛ = 10 : 1 supplies a 12 Ω load at 24 V. The primary current is
(A)2 A
(B)0.2 A
(C)20 A
(D)0.02 A
Show the solution
Given
Vₛ = 24 V, R = 12 Ω, Nₚ : Nₛ = 10 : 1
Asked
Iₚ
Concept
Find Iₛ, then use the current ratio.
Formula
Iₛ = Vₛ/R, Iₚ = Iₛ Nₛ/Nₚ
Baby steps
Iₛ = 24/12 = 2 A.
Iₚ = 2 × 1/10 = 0.2 A.
Answer
(B) 0.2 A
Why not the others
2 A is the secondary current. 20 A applies the ratio the wrong way. 0.02 A divides by 100.
Shortcut
Secondary first.
Where it went wrong
Ratio reversal.
Q16Numerical
A transformer of efficiency 95% delivers 1900 W. The power lost in it is
(A)95 W
(B)100 W
(C)1995 W
(D)5 W
Show the solution
Given
η = 0.95, Pₒᵤₜ = 1900 W
Asked
Loss
Concept
Pᵢₙ = Pₒᵤₜ/η; loss = Pᵢₙ − Pₒᵤₜ.
Formula
Pᵢₙ = Pₒᵤₜ/η
Baby steps
Pᵢₙ = 1900/0.95 = 2000 W.
Loss = 100 W.
Answer
(B) 100 W
Why not the others
95 W takes 5% of the output. 1995 W adds 95. 5 W uses the percentage as watts.
Shortcut
Find the input first.
Where it went wrong
Taking the loss as a percentage of the output.
Q17Concept
A transformer works on the principle of
(A)the magnetic effect of current alone
(B)self-induction
(C)mutual induction
(D)electrostatic induction
Show the solution
Given
Transformer
Asked
Principle
Concept
The primary's changing flux induces emf in the secondary.
Formula
—
Baby steps
Two coils, one induces emf in the other: mutual induction.
Answer
(C) mutual induction
Why not the others
Self-induction is a single coil. The magnetic effect alone is not induction. Electrostatic induction is about charges.
Shortcut
Two coils → mutual.
Where it went wrong
Choosing self-induction.
Q18Concept
A transformer is connected to a steady DC supply. After the switch is closed, the secondary voltage
(A)alternates at 50 Hz
(B)is steady and larger than the primary voltage in a step-up transformer
(C)is zero, apart from a brief pulse at switching
(D)equals the primary voltage
Show the solution
Given
DC input
Asked
Secondary output
Concept
Steady current → steady flux → no emf.
Formula
e = −N dφ/dt
Baby steps
The current rises briefly at switching: a short pulse.
Then the flux is steady: zero output.
Answer
(C) is zero, apart from a brief pulse at switching
Why not the others
A steady stepped-up output needs changing flux. A DC source cannot make 50 Hz. Equal voltage has no basis.
Shortcut
Transformers need AC.
Where it went wrong
Assuming the turns ratio still applies.
Q19Concept
The frequency of the output of a transformer
(A)is multiplied by the turns ratio
(B)equals the frequency of the input
(C)is divided by the turns ratio
(D)is always 50 Hz
Show the solution
Given
Frequency
Asked
Output frequency
Concept
The secondary emf follows the same alternating flux.
Formula
—
Baby steps
Same flux, same rate of alternation.
Frequency unchanged.
Answer
(B) equals the frequency of the input
Why not the others
The turns ratio changes voltage, not frequency. The output follows whatever the input frequency is.
Shortcut
Only V and I change.
Where it went wrong
Thinking a step-up raises frequency.
Q20Concept
A transformer cannot be called an amplifier because
(A)it has an iron core
(B)it cannot increase the voltage
(C)it cannot increase the power
(D)it works only on DC
Show the solution
Given
Amplifier
Asked
Why not
Concept
An amplifier gives power gain.
Formula
Pₒᵤₜ ≤ Pᵢₙ
Baby steps
A step-up transformer raises voltage but lowers current.
Power out is never more than power in.
Answer
(C) it cannot increase the power
Why not the others
It can raise voltage. The iron core is not the reason. It works on AC, not DC.
Shortcut
No power gain.
Where it went wrong
Thinking a voltage rise is amplification.
Q21Concept
In a step-up transformer, compared with the primary, the secondary has
(A)more turns, higher voltage and smaller current
(B)more turns, higher voltage and larger current
(C)fewer turns, higher voltage and smaller current
(D)more turns, lower voltage and larger current
Show the solution
Given
Step-up
Asked
Secondary compared with primary
Concept
V ∝ N, I ∝ 1/N.
Formula
Vₛ/Vₚ = Nₛ/Nₚ = Iₚ/Iₛ
Baby steps
More turns → more voltage.
Power balance → less current.
Answer
(A) more turns, higher voltage and smaller current
Why not the others
Larger current with higher voltage would create power. Fewer turns cannot step up. Lower voltage is step-down.
Shortcut
Voltage and current move opposite ways.
Where it went wrong
Letting current rise with voltage.
Q22Concept
The core of a transformer is made of soft iron because soft iron
(A)has very low permeability
(B)has high retentivity
(C)has high coercivity
(D)is easily magnetised and demagnetised, so hysteresis loss is small
Show the solution
Given
Core material
Asked
Reason
Concept
Low retentivity, low coercivity, high permeability.
Formula
—
Baby steps
The core is magnetised back and forth every cycle.
Soft iron does this with little energy lost.
Answer
(D) is easily magnetised and demagnetised, so hysteresis loss is small
Why not the others
High retentivity and high coercivity describe steel for permanent magnets. Low permeability would weaken coupling.
Shortcut
Soft iron = easy to reverse.
Where it went wrong
Mixing up soft iron and steel.
Q23Concept
In a step-down transformer, the secondary winding is made of thicker wire than the primary because
(A)it must withstand a higher voltage
(B)it has more turns
(C)it carries a larger current, and thicker wire reduces copper loss
(D)thick wire reduces eddy currents in the core
Show the solution
Given
Step-down windings
Asked
Reason for thicker wire
Concept
Copper loss I²R.
Formula
—
Baby steps
Step-down → larger secondary current.
Thicker wire → lower resistance → less I²R heat.
Answer
(C) it carries a larger current, and thicker wire reduces copper loss
Why not the others
Step-down secondaries have fewer turns and lower voltage. Wire thickness does not affect core eddy currents.
Shortcut
High-current coil → thick wire.
Where it went wrong
Linking thick wire to high voltage.
Q24Concept
Flux leakage in a transformer is reduced by
(A)winding both coils on a common soft iron core, one over the other
(B)using thicker wire
(C)laminating the core
(D)increasing the supply frequency
Show the solution
Given
Flux leakage
Asked
Remedy
Concept
Keep the flux in a closed iron path through both coils.
Formula
K → 1
Baby steps
A common core guides the flux through both coils.
Winding one over the other captures nearly all of it.
Answer
(A) winding both coils on a common soft iron core, one over the other
Why not the others
Thicker wire reduces copper loss. Lamination reduces eddy loss. Frequency does not stop leakage.
Shortcut
Match each loss to its remedy.
Where it went wrong
Using the lamination remedy for every loss.
Q25Concept
A transformer is usually more efficient than an electric motor of similar rating mainly because
(A)it has no copper windings
(B)it has no moving parts, so no mechanical losses
(C)it has no iron core
(D)it works on DC
Show the solution
Given
Efficiency comparison
Asked
Reason
Concept
Mechanical losses (friction) are absent.
Formula
—
Baby steps
Motors lose energy to friction and air resistance.
Transformers have no moving parts.
Answer
(B) it has no moving parts, so no mechanical losses
Why not the others
Transformers do have copper windings and iron cores, and they need AC.
Shortcut
No friction losses.
Where it went wrong
Guessing about materials.
Q26Graph
The primary voltage and primary turns of an ideal transformer are fixed. Which graph shows the secondary voltage against the number of secondary turns?
(A)
(B)
(C)
(D)
Show the solution
Given
Vₚ, Nₚ fixed
Asked
Vₛ against Nₛ
Concept
Vₛ ∝ Nₛ.
Formula
Vₛ = Vₚ Nₛ/Nₚ
Baby steps
Straight line through the origin.
Answer
(B) the graph in option B
Why not the others
The hyperbola is the current. The parabola squares Nₛ. The flat line ignores the turns.
Shortcut
Direct proportion.
Where it went wrong
Plotting the current graph.
Q27Graph
An ideal transformer delivers a fixed power. Which graph shows the secondary current against the number of secondary turns (primary fixed)?
(A)
(B)
(C)
(D)
Show the solution
Given
Fixed power
Asked
Iₛ against Nₛ
Concept
Iₛ ∝ 1/Nₛ.
Formula
Iₛ = Iₚ Nₚ/Nₛ
Baby steps
Inverse proportion: a hyperbola.
Answer
(D) the graph in option D
Why not the others
The rising line is the voltage. The flat line ignores the turns. A straight falling line would reach zero at a finite number of turns.
Shortcut
Inverse → hyperbola.
Where it went wrong
Ratio reversal on a graph.
Q28Graph
A fixed power is sent through a line of fixed resistance. Which graph shows the power lost in the line against the transmission voltage?
(A)
(B)
(C)
(D)
Show the solution
Given
Fixed P, R
Asked
Loss against V
Concept
Loss = P²R/V².
Formula
Loss ∝ 1/V²
Baby steps
Falls steeply at first, then flattens towards zero.
Answer
(B) the graph in option B
Why not the others
The rising graphs apply V²/R to the line wrongly. The flat line ignores the dependence on V.
Shortcut
Inverse square.
Where it went wrong
Using V²/R with the transmission voltage.
Q29Graph
A battery is connected to the primary of a transformer at t = 0 and left connected. Which graph shows the secondary voltage against time?
(A)
(B)
(C)
(D)
Show the solution
Given
DC switched on
Asked
Vₛ against t
Concept
Emf only while the primary current is changing.
Formula
Vₛ ∝ dIₚ/dt
Baby steps
Brief pulse while the current builds up.
Then zero.
Answer
(B) the graph in option B
Why not the others
A steady output needs changing flux. An oscillation needs AC. A ramp that stays up ignores that the flux stops changing.
Shortcut
Pulse, then nothing.
Where it went wrong
Assuming transformers work on DC.
Q30Assertion–reason
Assertion (A): A transformer cannot step up a DC voltage. Reason (R): A steady current in the primary produces no change of flux in the secondary.
(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.
Show the solution
Given
A: no DC operation. R: steady current, no changing flux.
Asked
Truth and link
Concept
Mutual induction needs changing flux.
Formula
e = −M dI/dt
Baby steps
A is true.
R is true.
R explains A.
Answer
(A) Both A and R are true, and R is the correct explanation of A.
Why not the others
(B) denies a direct link. (C) and (D) need a false statement.
Shortcut
Mechanism explains the limitation.
Where it went wrong
Choosing (B) by habit.
Q31Assertion–reason
Assertion (A): A step-up transformer increases the power supplied to a circuit. Reason (R): A step-up transformer increases the voltage.
(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.
Show the solution
Given
A: power up. R: voltage up.
Asked
Truth and link
Concept
Voltage rises, current falls, power at most equal.
Formula
Pₒᵤₜ ≤ Pᵢₙ
Baby steps
R is true.
A is false.
A false, R true.
Answer
(D) A is false, but R is true.
Why not the others
(A), (B) and (C) need A true.
Shortcut
Voltage gain ≠ power gain.
Where it went wrong
Equating voltage with power.
Q32Assertion–reason
Assertion (A): The output of a transformer has the same frequency as its input. Reason (R): A transformer has no moving parts.
(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.
Show the solution
Given
A: same frequency. R: no moving parts.
Asked
Truth and link
Concept
Frequency is unchanged because the secondary follows the same alternating flux.
Formula
—
Baby steps
A is true.
R is true.
Having no moving parts explains the lack of mechanical loss, not the frequency. R does not explain A.
Answer
(B) Both A and R are true, but R is not the correct explanation of A.
Why not the others
(A) claims a link that does not exist. (C) and (D) need a false statement.
Shortcut
Different facts, different reasons.
Where it went wrong
Linking two true transformer facts.
Q33Assertion–reason
Assertion (A): Electric power is transmitted over long distances at very high voltage. Reason (R): The power lost in a line is I²R, and for a given power a higher voltage means a smaller current.
(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.
Show the solution
Given
A: high-voltage transmission. R: I²R with smaller I.
Asked
Truth and link
Concept
Loss ∝ I² ∝ 1/V².
Formula
—
Baby steps
A is true.
R is true.
R explains A.
Answer
(A) Both A and R are true, and R is the correct explanation of A.
Why not the others
(B) denies a direct link. (C) and (D) need a false statement.
Shortcut
Loss reduction is the reason.
Where it went wrong
Choosing (B) out of caution.
Q34Assertion–reason
Assertion (A): Soft iron is used for transformer cores. Reason (R): Soft iron has high coercivity.
(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.
Show the solution
Given
A: soft iron cores. R: high coercivity.
Asked
Truth and link
Concept
Soft iron has low coercivity.
Formula
—
Baby steps
A is true.
R is false.
A true, R false.
Answer
(C) A is true, but R is false.
Why not the others
(A) and (B) need R true. (D) needs A false.
Shortcut
Soft = low coercivity.
Where it went wrong
Mixing up soft iron and steel.
Q35Two statements
Statement I: In an ideal transformer, Vₛ/Vₚ = Nₛ/Nₚ = Iₛ/Iₚ. Statement II: The efficiency of an ideal transformer is 100%.
(A)Both Statement I and Statement II are true.
(B)Both Statement I and Statement II are false.
(C)Statement I is true, but Statement II is false.
(D)Statement I is false, but Statement II is true.
Show the solution
Given
Ideal transformer relations
Asked
Which are true
Concept
Current ratio is inverse.
Formula
Iₛ/Iₚ = Nₚ/Nₛ
Baby steps
Statement I is false: Iₛ/Iₚ = Nₚ/Nₛ.
Statement II is true.
Answer
(D) Statement I is false, but Statement II is true.
Why not the others
Both-true fails on I. Both-false fails on II. I-true-II-false reverses them.
Shortcut
Voltage ratio up, current ratio down.
Where it went wrong
Ratio reversal hidden in a statement.
Answer key
1 D
2 C
3 C
4 A
5 A
6 D
7 A
8 D
9 A
10 A
11 C
12 D
13 B
14 C
15 B
16 B
17 C
18 C
19 B
20 C
21 A
22 D
23 C
24 A
25 B
26 B
27 D
28 B
29 B
30 A
31 D
32 B
33 A
34 C
35 D
Spread across letters: A 9, B 9, C 9, D 8. No letter repeats more than twice in a row. Question mix: Numerical 13, Concept 12, Graph 4, Assertion–reason 5, Two statements 1. Balancing seed 5000.