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Alternating Current — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsAlternating Current4 Marks
Question
Step-down transformers are used to decrease or step-down voltages. These are used when voltages need to be lowered for use in homes and factories. A small town with a demand of 800kW of electric power at 220V is situated 15km away from an electric plant generating power at 440V. The resistance of the two wire line carrying power is 0.5Q per km. The town gets power from the line through a 4000 - 220V step-down transformer at a sub-station in the town.
  1. The value of total resistance of the wires is:
  1. $25\Omega$
  2. $30\Omega$
  3. $35\Omega$
  4. $15\Omega$
  1. The line power loss in the form of heat is:
  1. 550kW
  2. 650kW
  3. 600kW
  4. 700kW
  1. How much power must the plant supply, assuming there is negligible power loss due to leakage?
  1. 600kW
  2. 1600kW
  3. 500W
  4. 1400kW
  1. The voltage drop in the power line is:
  1. 1700V
  2. 3000V
  3. 2000V
  4. 2800V
  1. The total value of voltage transmitted from the plant is:
  1. 500V
  2. 4000V
  3. 3000V
  4. 7000V
✓

Answer

  1. (d) $15\Omega$
Explanation:
Resistance of the two wire lines carrying power $=0.5\frac{\Omega}{\text{Km}}$
Total resistance $=(15+15)0.5=15\Omega$
  1. (c) 600kW
Explanation:
Line power loss $= I^2R$
RMS current in the coil,
$\text{I}=\frac{\text{P}}{\text{V}_1}=\frac{800\times10^3}{4000}=200\text{A}$
$\therefore$ Power loss $= (200)^2 \times 15 = 600kW$
  1. (d) 1400kW
Explanation:
Assuming that the power loss is negligible due to the leakage of the current.
The total power supplied by the plant,
= 800kW + 600kW = 1400kW
  1. (b) 3000V
Explanation:
Voltage drop in the power line = IR
= 200 × 15 = 3000V
  1. (d) 7000V
Explanation:
Total voltage transmitted from the plant,
= 3000V + 4000V = 7000V

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