A transformer consisting of \(300\) turns in the primary and \(150\) turns in the secondary gives output power of \(2.2\, kW\). If the current in the secondary coils is \(10\, A\), then the input voltage and current in the primary coil are

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Answer & Solution

Correct Answer

(A) \(440\, V\) and \(5\, A\)

Detailed explanation

Given \(\mathrm{N}_{\mathrm{P}}=300, \mathrm{N}_{\mathrm{s}}=150, \mathrm{P}_{0}=2200 \,\mathrm{W}\) \(\mathrm{l}_{\mathrm{s}}=10\, \mathrm{A}\)

\(\mathrm{P}_{0}=\mathrm{V}_{0} \mathrm{l}_{0} \)

\(\Rightarrow 2200=\mathrm{V}_{0} \times 10 \)

\(\Rightarrow \mathrm{V}_{0}=220\, \mathrm{V}\)

\(\because \frac{V_{i}}{V_{0}}=\frac{N_{p}}{N_{s}} \)

\(\Rightarrow V_{1}=2 \times 220=440\, \mathrm{V}\)

Also, \(\quad \mathrm{P}_{0}=\mathrm{V}_{\mathrm{i}} \mathrm{l}_{\mathrm{i}}\)

\(\Rightarrow \mathrm{I}_{\mathrm{i}}=\frac{2200}{440}=5\, \mathrm{A}\)

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List-I (Relation)		List-II (Law)
A.	\(\oint \overrightarrow{ E } \cdot \overrightarrow{ d } l =-\frac{ d }{ dt } \oint \overrightarrow{ B } \cdot \overrightarrow{ d a}\)	I.	Ampere's circuital law.
B.	\(\oint\) B. \(\vec{d} l=\mu_0\left(1+\epsilon_0 \frac{d \phi_E}{d t}\right)\)	II.	Faraday's laws of electromagnetic induction.
C.	\(\oint \overrightarrow{ E } . \overrightarrow{ d a}=\frac{1}{\epsilon_0} \int \rho dv\)	III.	Ampere-Maxwell law
D.	\(\oint \overrightarrow{ B } \cdot \overrightarrow{ d } l =\mu_0 I\)	IV.	Gauss's law of electrostatics