Transformers

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   Laminated iron sheets are used to minimize eddy currents in the core of a transformer (lamination raises the effective resistance to the circulating eddy currents, cutting the associated energy loss).

2. 2.↑ back to the question

   (i) For $\omega<\omega_0$, $X_C=\dfrac1{\omega C}>X_L=\omega L$, so $X_C>X_L$. Since $\tan\phi=\dfrac{X_L-X_C}{R}<0$, $\phi$ is negative, meaning the current leads the voltage (the circuit behaves as capacitor-dominated). (ii) A transformer works only on mutual induction, which requires a continuously changing magnetic flux. A steady dc voltage on the primary produces a constant (unchanging) flux, so no emf is induced in the secondary — hence a transformer cannot step up (or step down) a dc voltage. (iii) A metal detector works on the principle of electromagnetic induction: an ac-driven coil near a metal object induces eddy currents in the metal, which alter the effective inductance/impedance seen by the detector's circuit, triggering detection.

3. 3.↑ back to the question

   (a) For an ideal transformer, $P_{in}=P_{out}\Rightarrow I_pV_p=I_sV_s$. Since $V_s>V_p$ (step-up), $\dfrac{I_p}{I_s}=\dfrac{V_s}{V_p}>1$, so the output (secondary) current $I_s$ is less than the input (primary) current $I_p$. (b) Laminating the iron core cuts it into thin insulated sheets, reducing induced eddy currents and hence the associated heating losses. (c) A real transformer always has some energy losses — copper losses (resistive heating in the windings), hysteresis losses, eddy-current losses and flux-leakage losses — so the output power is always somewhat less than the input power.

4. 4.↑ back to the question

   A step-down transformer has a primary coil of $N_p$ turns and a secondary coil of $N_s$ turns ($N_s<N_p$), both wound on a common soft-iron core. An alternating $V_p$ on the primary drives an alternating current, setting up a changing flux $\phi$; with primary resistance small, $V_p\approx\varepsilon_p=N_p\dfrac{d\phi}{dt}$. The same flux links the secondary via the core, inducing $\varepsilon_s=N_s\dfrac{d\phi}{dt}$; with secondary open, $V_s=\varepsilon_s$. $\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}<1 \Rightarrow V_s<V_p$ — the output voltage is stepped down. The core is laminated (thin, mutually insulated iron sheets) to minimize eddy currents induced by the changing flux, reducing heating loss and increasing efficiency.

5. 5.↑ back to the question

   A step-up transformer has a primary coil of $N_p$ turns and secondary of $N_s$ turns ($N_s>N_p$), both wound on a common soft-iron core. An alternating $V_p$ on the primary produces a changing flux $\phi$, with $V_p\approx\varepsilon_p=N_p\dfrac{d\phi}{dt}$ (primary resistance small). The soft-iron core links this flux to the secondary, inducing $\varepsilon_s=N_s\dfrac{d\phi}{dt}=V_s$ (secondary open): $\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}>1$, so the output is stepped up. For an ideal transformer, $I_sV_s=I_pV_p \Rightarrow \dfrac{I_s}{I_p}=\dfrac{N_p}{N_s}$. (i) The core is laminated to minimize eddy currents (and the $I^2R$ heating they cause), by restricting the paths available to these circulating currents. (ii) Thick copper wire has low resistance, minimizing $I^2R$ (copper/Joule heating) losses in the windings.

6. 6.↑ back to the question

   (a) The device is a step-down transformer, which works on mutual induction — a changing primary current produces a changing flux which, via a common soft-iron core, induces an emf in the secondary coil. Four sources of energy loss: (i) copper losses ($I^2R$ heating in winding resistance); (ii) hysteresis losses (repeated magnetization/demagnetization of the core each cycle); (iii) eddy-current (iron) losses in the core (minimized by lamination); (iv) flux-leakage losses (imperfect coupling between the coils). (b) $P=1200\text{ kW}=1200\times10^3$ W. Line resistance $R=2\times20\times0.5=20\ \Omega$ (20 km each way). Transmission-line voltage, $E_v=4000$ V. $I_v=\dfrac{P}{E_v}=\dfrac{1200\times10^3}{4000}=300\text{ A}$ Line power loss $=I_v^2R=(300)^2\times20=1.8\times10^6\text{ W}=1800\text{ kW}$

7. 7.↑ back to the question

   (a) Diagram: primary coil ($N_p$ turns) and secondary coil ($N_s$ turns) wound on a common soft-iron core, primary connected to the ac input, secondary connected to the load. $V_p\approx\varepsilon_p=-N_p\dfrac{d\phi}{dt}$; $V_s=\varepsilon_s=-N_s\dfrac{d\phi}{dt}$ (secondary open) $\Rightarrow \dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}$ For an ideal transformer, $I_sV_s=I_pV_p \Rightarrow \dfrac{V_s}{V_p}=\dfrac{I_p}{I_s}=\dfrac{N_s}{N_p}$ (b) $N_p=3000$, $V_p=2200$ V, $V_s=220$ V. $N_s=\dfrac{N_pV_s}{V_p}=\dfrac{3000\times220}{2200}=300$

8. 8.↑ back to the question

   (i) Diagram: primary coil ($N_p$ turns) and secondary coil ($N_s$ turns, $N_s<N_p$) wound on a common laminated soft-iron core. Principle: mutual induction — a changing current in the primary induces an emf in the secondary via the changing flux linked through the core. (ii) $\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}$ (iii) $I_pV_p=I_sV_s \Rightarrow \dfrac{I_p}{I_s}=\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}$ (iv) $P_{in}=P_{out}=550$ W, $V_p=220$ V. $I_p=\dfrac{P_{in}}{V_p}=\dfrac{550}{220}=2.5\text{ A}$

9. 9.↑ back to the question

   (i) A transformer changes (steps up or down) an alternating voltage; it works on mutual induction — an alternating primary current produces a changing flux, which, through a common soft-iron core, induces an emf in the secondary coil. Energy losses: copper losses ($I^2R$ heating in the windings), hysteresis losses (in the core), eddy-current losses (minimized by lamination), and flux-leakage losses. (ii) Given $N_p=100$, $\dfrac{N_s}{N_p}=100$, $\varepsilon_i=\varepsilon_p=220$ V, $P_i=1100$ W. (a) $N_s=100\times N_p=100\times100=10000$ (b) $I_p=\dfrac{P_i}{\varepsilon_p}=\dfrac{1100}{220}=5\text{ A}$ (c) $\varepsilon_s=\dfrac{N_s}{N_p}\times\varepsilon_p=100\times220=22000\text{ V}$ (d) For an ideal transformer, $P_o=P_i=1100$ W $\Rightarrow I_s=\dfrac{P_o}{\varepsilon_s}=\dfrac{1100}{22000}=0.05\text{ A}$ (e) $P_s=P_o=P_i=1100\text{ W}$

10. 10.↑ back to the question

    (a) Diagram: primary coil ($N_p$ turns) wound directly over (concentric with) the secondary coil ($N_s$ turns), both on a common soft-iron core, minimizing flux leakage. (b) Principle: mutual induction — an alternating primary current produces a continuously changing flux, which, linked with the secondary through the soft-iron core, induces an alternating emf in it. With primary resistance negligible, $V_p\approx\varepsilon_p=-N_p\dfrac{d\phi}{dt}$; with secondary open, $V_s=\varepsilon_s=-N_s\dfrac{d\phi}{dt}$. (i) $\dfrac{V_s}{V_p}=\dfrac{N_s}{N_p}$ (ii) For an ideal transformer, $I_sV_s=I_pV_p \Rightarrow \dfrac{V_s}{V_p}=\dfrac{I_p}{I_s}=\dfrac{N_s}{N_p}$ (c) Main assumptions: the resistance of the primary and secondary windings is negligible, and the same flux links both windings (no flux leakage from the core). (d) Two reasons: (i) copper losses — $I^2R$ heating in the winding resistance; (ii) hysteresis losses — due to repeated magnetization/demagnetization of the core each cycle. [Also acceptable: eddy-current losses, flux-leakage losses.]

11. 11.Answer: (D)↑ back to the question

    Lamination cuts the core into thin, mutually insulated sheets, which restricts the paths available for induced eddy currents, reducing eddy-current loss. Hence option (d).