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Alternating Current — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsAlternating Current4 Marks
Question
A transformer is essentially an $a.c.$ device. It cannot work on $d.c.$ It changes altemating voltages or currents. It does not affect the frequency of $a.c.$ It is based on the phenomenon of mutual induction. A transformer essentially consists of two coils of insulated copper wire having different number of tums and wound on the same soft iron core.
The number of turns in the primary and secondary coils of an ideal transfonner are $2000$ and $50$ respectively. The primary coil is connected to a main supply of $120V$ and secondary coil is connected to a bulb of resistance $0.6\Omega.$
  1. The value of voltage across the secondary coil is:
  1. $5V$
  2. $2V$
  3. $3V$
  4. $10V$
  1. The value of current in the bulb is:
  1. $7A$
  2. $15A$
  3. $3A$
  4. $5A$
  1. The value of current in primary coil is:
  1. $0.125A$
  2. $2.52A$
  3. $1.51A$
  4. $3.52A$
  1. Power in primary coil is:
  1. $20W$
  2. $5W$
  3. $10W$
  4. $15W$
  1. Power in secondary coil is:
  1. $15W$
  2. $20W$
  3. $7W$
  4. $8W$
✓

Answer

  1. $(c) \ 3V$
As, $\frac{\text{E}_\text{s}}{\text{E}_\text{p}}=\frac{\text{n}_\text{s}}{\text{n}_\text{p}}$
$\Rightarrow\text{E}_\text{s}=\text{E}_\text{p}\times\frac{\text{n}_\text{p}}{\text{n}_\text{p}}$
$=\frac{120\times50}{2000}$
$=3\text{V}$
  1. $(d) \ 5A$
$\text{I}_0=\frac{\text{E}_\text{S}}{\text{R}}$
$\Rightarrow\text{I}_0=\frac{3}{0.6}=5\text{A}$
  1. $(a) \ 0.125A$
As, $\frac{\text{I}_\text{P}}{\text{I}_\text{S}}=\frac{\text{E}_\text{S}}{\text{E}_\text{P}}$
$\Rightarrow\text{I}_\text{p}=\frac{\text{E}_\text{S}}{\text{E}_\text{P}}\times\text{I}_\text{S}$
$=\frac{3}{120}\times5$
$=0.125\text{A}$
  1. $(d) \ 15W$
Power in primary, $P_{P }$
$= E_P \times I_{P }$
$= 120 \times 0.125$
$= 15W$
  1. $(a) \ 15W$
Power in secondary coil, $P_S = E_S \times I_S $
​​​​​​​$= 3 \times 5$
$= 15W$

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