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MODEL PAPER 7 — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsMODEL PAPER 75 Marks
Question
$a$. Draw the diagram of a device which is used to decrease high $ac$ voltage into a low $ac$ voltage and state its working principle. Write four sources of energy loss in this device.
$b$. A small town with a demand of $1200 kW$ of electric power at $220 V$ is situated $20 \ km$ away from an electric plant generating power at $440 V$ . The resistance of the two wire line carrying power is $0.5 \Omega \ \text{per} \ \ km$. The town gets the power from the line through a $4000-220 V$ step $-$ down transformer at a sub $-$ station in the town. Estimate the line power loss in the form of heat
✓

Answer

$a$. The device used to decrease high $ac$ voltage into a low $ac$ voltage is called transformer $($step $-$ down transformer$)$.
Working principle:
Transformer works on the principle of Faraday’s law of electromagnetic induction.
The law of electromagnetic induction states that when magnetic flux linked with a coil changes, an emf is induced in the coil.
Transformer consists of two coils called primary coil and secondary coil.
The ac current in primary coil changes magnetic flux linked with the secondary coil and thus an emf is induced in the secondary coil.
Image

Sources of energy loss in transformer
$i$. Copper loss: The coils of transformer $($made of copper$)$ have a finite resistance due to which some energy in lost as heat.
$ii$. Iron loss: Due to induced eddy currents in the iron core, some energy is lost in the bulk.
$iii.$ Magnetic loss: Since all magnetic flux in primary coil does not pass through the secondary coil, there is some loss of energy due to leakage of flux.
$iv$. Hysteresis loss: alternating magnetization and demagnetization of the iron core cause some loss of energy in form of heat
$b$. Demand of electric power $= 1200 kW$
Distance of town from power station $= 20 \ km$ Two wire $= 20 \times 2 = 40 \ km$
Total resistance of line $= 40 \times 0.5 = 200$
The town gets a power of $4000$ volts
Power $=$ voltage current
$I=\frac{1200 \times 10^3}{4000}=\frac{1200}{4}=300 A$
The line power loss in the form of heat $= I ^2 \times R$
$=(300)^2 \times 2$
$=9000 \times 20=1800 kW$

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