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Alternating Current — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsAlternating Current4 Marks
Question
A transformer is an electrical device which is used for changing the a.c. voltages. It is based on the phenomenon of mutual induction i.e. whenever the amount of magnetic flux linked with a coil changes, an e.m.f. is induced in the neighbouring coil. For an ideal transformer, the resistances of the primary and secondary windings are negligible.

It can be shown that $\frac{\text{E}_\text{S}}{\text{E}_\text{P}}=\frac{\text{I}_\text{P}}{\text{I}_\text{S}}=\frac{\text{n}_\text{S}}{\text{n}_\text{P}}=\text{K}$
where the symbols have their standard meanings.
For a step up transformer, $\text{n}_\text{S} > \text{n}_\text{P}; \text{E}_\text{S} > \text{E}_\text{P}; \text{k} > \text{I}; \therefore \text{I}_\text{S} < \text{I}_\text{P}$
For a step down transformer, $\text{n}_\text{S} > \text{n}_\text{P}; \text{E}_\text{S} > \text{E}_\text{P}; \text{k} > \text{I};$
The above relations are on the assumptions that efficiency of transfonner is 100%.
lnfact, efficiency $\eta=\frac{\text{output power}}{\text{intput power }}=\frac{\text{E}_\text{S}\text{I}_\text{S}}{\text{E}_\text{P}\text{I}_\text{P}}$
  1. Which of the following quantity remains constant in an ideal transformer?
  1. Current.
  2. Voltage.
  3. Power.
  4. All of these.
  1. Transformer is used to.
  1. Convert ac to de voltage.
  2. Convert de to ac voltage.
  3. Obtain desired de power.
  4. Obtain desired ac voltage and current.
  1. The number of tums in primary coil of a transformer is 20 and the number of turns in a secondary is 10. If the voltage across the primary is ac 220V, what is the voltage across the secondary?
  1. ac 100V
  2. ac 120V
  3. ac 110V
  4. ac 220V
  1. In a transformer the number of primary turns is four times that of the secondary turns. Its primary is connected to an a.c. source of voltage V. Then,
  1. Current through its secondary is about four times that of the current through its primary.
  2. Voltage across its secondary is about four times that of the voltage across its primary.
  3. Voltage across its secondary is about two times that of the voltage across its primary.
  4. voltage across its secondary is about $\frac{1}{2\sqrt{2}}$ times that of the voltage across its primary.
  1. A transformer is used to light 100W-110V lamp from 220V mains. If the main current is 0.5A, the efficiency of the transformer is:
  1. 95%
  2. 99%
  3. 90%
  4. 96%
✓

Answer

  1. (c) Power
Explanation:
In an ideal transformer, there is no power loss. The efficiency of an ideal transformer is $\eta=1$ (i.e 100%) i.e. input power = output power.
  1. (d) Obtain desired ac voltage and current.
Explanation:
Transformer is used to obtain desired ac voltage and current.
  1. (c) ac 110V
Explanation:
For a transformer, $\frac{\text{V}_\text{S}}{\text{V}_\text{p}}=\frac{\text{n}_\text{S}}{\text{n}_\text{p}}$
Where N denotes number of turns and V = voltage.
$\therefore{\text{V}_\text{S}}={\text{ac}\ 110}{\text{V}}$
  1. (a) Current through its secondary is about four times that of the current through its primary.
Explanation:
In a transformer the primary and secondary currents are related by,
$\text{I}_\text{S}=\Big(\frac{\text{N}_\text{S}}{\text{N}_\text{P}}\Big)\text{I}_\text{P}$
And the Voltage are related by,
$\text{V}_\text{S}=\Big(\frac{\text{N}_\text{P}}{\text{N}_\text{S}}\Big)\text{V}_\text{P}$
where subscripts p and s refer to the primary and secondary of the transformer.
Here, $\text{V}_\text{P}=\text{V},\frac{\text{N}_\text{P}}{\text{N}_\text{S}}=4\ \therefore\text{I}_\text{P}=4\text{I}_\text{P}$
and, $\text{V}_\text{S}=\Big(\frac{1}{4}\Big)\text{V}=\frac{\text{V}}{4}$
  1. (c) 90%
Explanation:
The efficiency of the transformer is:
$\eta=\frac{\text{output power}(\text{p}_\text{out})}{\text{intput power}(\text{p}_\text{in})}\times100$
Here, $\mathrm{P_{out}= 100W, P_{in}= (220V)(0.5A) = 110W}$
$\therefore\eta=\frac{100\text{W}}{110\text{W}}\times100=90\%$

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