KCET · Physics · Alternating Current
For a transformed, the turns ratio is 3 and its efficiency is \(0.75\). The current flowing in the primary coil is \(2 \mathrm{~A}\) and the voltage applied to it is \(100 \mathrm{~V}\). Then the voltage and the current flowing in the secondary coil are ... respectively
- A \(150 \mathrm{~V}, 1.5 \mathrm{~A}\)
- B \(300 \mathrm{~V}, 0.5 \mathrm{~A}\)
- C \(300 \mathrm{~V}, 1.5 \mathrm{~A}\)
- D \(150 \mathrm{~V}, 0.5 \mathrm{~A}\)
Answer & Solution
Correct Answer
(B) \(300 \mathrm{~V}, 0.5 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
Given, \(\frac{n_{s}}{n_{P}}=3 ; \quad \eta=0.75 ; I_{P}=2 \mathrm{~A}\) and \(V_{P}=100 \mathrm{~V}\)
We know that,
\(\eta=\frac{V_{S}}{V_{P}} \frac{I_{S}}{I_{P}} \)
\( \text { or } \eta=\left(\frac{n_{s}}{n_{P}}\right) \frac{I_{s}}{I_{P}} \left(\because \frac{n_{s}}{n_{p}}=\frac{V_{s}}{V_{p}}\right) \)
\( \Rightarrow 0.75=\frac{3 \times I_{s}}{2} \)
\( \Rightarrow I_{s}=0.5 \mathrm{~A} \)
By the relation, \(\frac{n_{s}}{n_{P}}=\frac{V_{s}}{V_{P}} \)
\( \Rightarrow 3=\frac{V_{s}}{100} \Rightarrow V_{s}=300 \mathrm{~V}\)
We know that,
\(\eta=\frac{V_{S}}{V_{P}} \frac{I_{S}}{I_{P}} \)
\( \text { or } \eta=\left(\frac{n_{s}}{n_{P}}\right) \frac{I_{s}}{I_{P}} \left(\because \frac{n_{s}}{n_{p}}=\frac{V_{s}}{V_{p}}\right) \)
\( \Rightarrow 0.75=\frac{3 \times I_{s}}{2} \)
\( \Rightarrow I_{s}=0.5 \mathrm{~A} \)
By the relation, \(\frac{n_{s}}{n_{P}}=\frac{V_{s}}{V_{P}} \)
\( \Rightarrow 3=\frac{V_{s}}{100} \Rightarrow V_{s}=300 \mathrm{~V}\)
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