Solution:

Ans. Try yourself, Similar to Q. 1 (i), Long Answer type Questions.

Two reasons for energy losses

Flux leakage / joule heat losses in the windings / Eddy currents / hysteresis

(Any two) $1 / 2+1 / 2$

[CBSE Marking Scheme 2016]

[AT Q. 7. (i) Write the function of a transformer. State its principle of working with the help of a diagram. Mention various energy losses in this device.

(ii) The primary coil of an ideal step up transformer has 100 turns and transformation ratio is also 100. The input voltage and power are respectively $220 \mathrm{~V}$ and $1100 \mathrm{~W}$. Calculate

(a) number of turns in secondary

(b) current in primary

(c) voltage across secondary

(d) current in secondary

(e) power in secondary

R [Delhi I, II, III 2016]

Ans. (i) Conversion of $a c$ of low voltage into $a c$ of high voltage & vice versa.

Working Principle :

Mutual induction : When alternating voltage is applied to primary windings, emf is induced in the secondary windings.

(a)

(b) (Any one of the above diagram)

Energy losses :

• Leakage of magnetic flux
• Eddy currents
• Hysteresis loss
• Copper loss

(ii)

$$ N_{p}=100 $$

Transformation ratio $=100$

(a) Number of turns in secondary coil

$$ N_{S}=100 \times 100=10000 $$

(b) Input Power $=$ Input voltage $\times$ current in primary

$$ I_{p}=5 \mathrm{~A} $$

(c)

$$ \frac{V_{S}}{V_{P}}=\frac{N_{S}}{N_{P}} $$

$$ \frac{V_{S}}{220}=100 $$

$$ \Rightarrow \quad V_{S}=2.2 \times 10^{4} \text { volts } $$

(d)

$$ \frac{I_{P}}{I_{S}}=\frac{N_{S}}{N_{P}} $$

(e) Power in secondary $=$ Power in Primary

- $1100 \mathrm{~W}$

[CBSE Marking Scheme 2016]