Q.Explain copper-loss in a transformer.

Showing the 12 most recent of 17 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.An ideal transformer has 500 turns in the primary and 5000 turns in the secondary. If the primary be connected to a 6 V battery, then the secondary voltage is

   (a) 0

   (b) 0.6 V

   (c) 60 V

   (d) 6 V

   ›Reveal solutionSolution

   A transformer needs a changing current/flux to work. A DC battery gives a constant current, so once steady state is reached the secondary voltage is 0.

   A transformer works on the principle of mutual induction: the emf induced in the secondary is

   $e_s = -M\dfrac{di_p}{dt}$

   …

   🔒 Start your 14-day free trial to see the full solution →
2. CBSE 2026Set ANNUAL1 mark
   Q.Why cannot a transformer be used to step up direct current (D.C.)?

   ›Reveal solutionSolution

   No changing flux, no induced EMF — a transformer needs AC to work at all.

   A transformer operates on the principle of mutual induction: a time-varying current in the primary coil produces a time-varying magnetic flux in the core, which links the secondary coil and induces an EMF in it, given by $\varepsilon_2 = -N_2\dfrac{d\Phi}{dt}$. With a constant DC current in the primary, the flux in the core, once established, remains steady (constant) — its rate of change $d\Phi/dt$ is zero in the steady state. Since the induced EMF depends entirely on this rate of change, no EMF (and hence no stepped-up voltage) is induced in the secondary for steady DC, so a t …

   🔒 Start your 14-day free trial to see the full solution →
3. CBSE 2025Set X11 markMCQ
   Q.Transformer cores are usually laminated. This is to reduce energy loss due to

   (a) flux leakage

   (b) winding resistance

   (c) eddy currents

   (d) hysteresis

   ›Reveal solutionSolution

   (c) eddy currents. The changing flux in the core induces circulating (eddy) currents in the solid metal, which dissipate energy as heat ($\propto$ resistance path). Laminating the core with thin i …

   🔒 Start your 14-day free trial to see the full solution →
4. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which quantity is increased in a step-up transformer ?

   (a) current

   (b) voltage

   (c) power

   (d) frequency

   ›Reveal solutionSolution

   A step-up transformer increases voltage (and correspondingly decreases current), since power and frequency stay the same.

   For an ideal transformer,

   $\frac{V_s}{V_p} = \frac{N_s}{N_p}$ …

   🔒 Start your 14-day free trial to see the full solution →
5. CBSE 2025Set ANNUAL1 mark
   Q.On which principle does transformer work?

   ›Reveal solutionSolution

   A transformer transfers energy from primary to secondary coil through mutual induction of a changing magnetic flux.

   A transformer consists of two coils (primary and secondary) wound on a common laminated soft-iron core. When an alternating current is passed through the primary coil, it produces a continuously changing magnetic flux in the core. Since the secondary coil is linked to the same core, this changing flux also links the secondary coil.

   By Faraday's law of electromagnetic induction, a changing flux linked with the secondary coil induces an alternating emf in it — this is mutual induction, i.e., induction of emf in one coil due to a changing current in a nearby (magnetically coupled) coil.

   …

   🔒 Start your 14-day free trial to see the full solution →
6. CBSE 2025Set ANNUAL1 mark
   Q.Why is electric power transmission from power stations to sub-stations near consumers done at high voltages ?

   ›Reveal solutionSolution

   For a fixed power to be delivered, $P=VI$, so raising the transmission voltage lowers the current; since resistive line loss goes as $I^2R$, a lower current means far less energy is wasted as heat.

   Electrical power transmitted is $P = VI$. For a given power $P$ to be delivered by the transmission line, increasing the transmission voltage $V$ proportionally decreases the current $I = P/V$.

   The power dissipated as heat in the transmission line's resistance $R$ is

   $P_{loss} = I^2 R$

   …

   🔒 Start your 14-day free trial to see the full solution →
7. CBSE 2024Set 55/2/11 markMCQ
   Q.Which of the following quantity/quantities remains same in primary and secondary coils of an ideal transformer ? Current, Voltage, Power, Magnetic flux (A) Current only (B) Voltage only (C) Power only (D) Magnetic flux and Power both

   ›Reveal solutionSolution

   In an ideal transformer, the magnetic flux linking both coils is the same (by Faraday’s law), and power is conserved (no losses). Current and voltage change with the turns ratio. So the correct choice is (D) Magnetic flux and Power both.

   ---

   The core idea

   An ideal transformer is a perfect magnetic circuit with no energy losses — no resistance in the windings, no hysteresis, no eddy currents, and perfect coupling (all flux from the primary passes through the secondary).

   Two fundamental principles govern it:

   1. Faraday’s law of induction — the same changing magnetic flux $\Phi$ links every turn of both coils.
   2. Conservation of energy — in the absence of losses, the power delivered to the primary must equal the power extracted from the secondary.

   From these, everything else follows.

   ---

   Step-by-step reasoning

   1. Magnetic flux is the same in both coils In an ideal transformer, the core is assumed to have zero reluctance and no flux leakage. The alternating current in the primary creates a time-varying magnetic flux $\Phi(t)$ that is entirely confined to the core. Since both coils are wound on the same core, every turn of the secondary is linked by exactly the same flux as every turn of the primary. By Faraday’s law, the induced emf in each coil is proportional to the number of turns:

   $\mathcal{E}_p = -N_p \frac{d\Phi}{dt}, \qquad \mathcal{E}_s = -N_s \frac{d\Phi}{dt}$

   The flux $\Phi$ itself is identical — only the induced voltages differ because $N_p \neq N_s$.

   2. Voltage changes with the turns ratio From the above,

   $\frac{V_s}{V_p} = \frac{N_s}{N_p}$

   So voltage is not the same in primary and secondary unless $N_p = N_s$ (which is not generally true).

   3. Current changes inversely with the turns ratio For an ideal transformer, the magnetising current is negligible, and the primary current adjusts to balance the secondary load. Power conservation gives:

   $V_p I_p = V_s I_s \quad \Rightarrow \quad \frac{I_s}{I_p} = \frac{N_p}{N_s}$

   So current is also not the same.

   4. Power is conserved …

   🔒 Start your 14-day free trial to see the full solution →
8. CBSE 2024Set IMPROVEMENT1 markMCQ
   Q.In actual transformer, reason of energy losses is —

   (a) Flux Leakage

   (b) Eddy Currents

   (c) Resistance of the windings

   (d) All of above

   ›Reveal solutionSolution

   A real transformer loses energy through flux leakage, eddy currents, AND winding resistance — all three act together.

   An ideal transformer has no losses, but an actual transformer loses energy due to: (1) Flux leakage — not all of the flux produced by the primary links the secondary. (2) Eddy currents induced in the iron core, which dissipate energy as heat (reduced, but not eliminated, by lamination). (3) Resistance of the copper windin …

   🔒 Start your 14-day free trial to see the full solution →
9. CBSE 2024Set ANNUAL1 markMCQ
   Q.The core in transformers and other electromagnetic devices is laminated, so as to

   (a) increase the magnetic field

   (b) increase the magnetic flux

   (c) reduce the magnetism in the core

   (d) reduce the eddy current losses in the core

   ›Reveal solutionSolution

   A changing magnetic flux through any conducting core induces circulating "eddy" currents in the body of the core itself; these currents dissipate energy as heat ($I^2R$) without doing any useful work. Laminating the core cuts this loss.

   Why the core is laminated

   In a transformer (or any a.c. electromagnetic device), the iron core carries a time-varying flux $\Phi(t)$. By Faraday's law this changing flux induces an e.m.f. not only in the windings but also within the bulk of the iron itself, driving circulating eddy currents inside the core.

   • If the core were a single solid block, these eddy currents could flow in large loops of large cross-sectional area, and the power dissipated by eddy currents scales as $P_{eddy} \propto B_{max}^2\, f^2\, t^2$ where $t$ is the thickness of the conducting slab carrying the loop. …

   🔒 Start your 14-day free trial to see the full solution →
10. CBSE 2022Set HE2171 markMCQ
    Q.The transformer is based on the principle:

    (i) Self Induction

    (ii) Mutual Induction

    (iii) Electromagnetic waves

    (iv) None of the above

    ›Reveal solutionSolution

    A transformer works on the principle of mutual induction between its primary and secondary windings.

    A transformer consists of two coils (primary and secondary) wound on a common laminated iron core. When an alternating current flows in the primary, it produces a continuously changing magnetic flux in the core; this changing flux links the secondary coil as well and, by Faraday's law, induces an alternating emf in it. This transfer of energy from one coil to another purely through a …

    🔒 Start your 14-day free trial to see the full solution →
11. CBSE 2022Set TERM21 markMCQ
    Q.Transformer is based upon the principle of:

    (a) Self-induction

    (b) Mutual induction

    (c) Eddy current

    (d) Angular momentum

    ›Reveal solutionSolution

    A transformer transfers energy between its primary and secondary coils purely through mutual electromagnetic induction.

    A transformer consists of two coils (primary and secondary) wound on a common laminated iron core. An alternating current in the primary coil produces a changing magnetic flux in the core; this changing flux links the secondary coil and, by Faraday's law, induces an emf in it. This transfer of energy fro …

    🔒 Start your 14-day free trial to see the full solution →
12. CBSE 2022Set ANNUAL1 markMCQ
    Q.A step-down transformer reduces the supply voltage from 220 V to 11 V and increases the current from 6 A to 100 A. Then its efficiency is :

    (a) 0.12

    (b) 1.2

    (c) 0.9

    (d) 0.83

    ›Reveal solutionSolution

    Dividing output power ($11\text{ V}\times100\text{ A}$) by input power ($220\text{ V}\times6\text{ A}$) gives a transformer efficiency of about 0.83.

    Working

    Efficiency of a transformer is defined as

    $\eta=\dfrac{\text{Output power}}{\text{Input power}}$

    Input (primary) power:

    $P_{in}=V_pI_p=220\times6=1320\ \text{W}$

    …

    🔒 Start your 14-day free trial to see the full solution →