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Alternating Current — Physics STD 12 Science — Question

Rajasthan 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 $15\ km$ 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. $(d) 15\Omega$
Resistance of the two wire lines carrying power $=0.5\frac{\Omega}{\text{ Km}}$
Total resistance $=(15+15)0.5=15\Omega$
  1. $(c)\ 600kW$
Line power loss $= I^2R$
$\text{RMS}$ 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. $(d) \ 1400kW$
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. $(b)\ 3000V$
Voltage drop in the power line $= IR$
$= 200 \times 15 = 3000V$
  1. $(d) \ 7000V$
Total voltage transmitted from the plant,
$= 3000V + 4000V = 7000V$

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