MHT CET · Physics · Alternating Current
The primary and secondary voltage of an ideal step-down transformer is \(200 \mathrm{~V}\) and \(25 \mathrm{~V}\) respectively. The secondary is connected to a device, which draws a current of \(2 \mathrm{~A}\). The current in the primary is
- A \(25 \mathrm{~mA}\)
- B \(42 \mathrm{~mA}\)
- C \(160 \mathrm{~mA}\)
- D \(250 \mathrm{~mA}\)
Answer & Solution
Correct Answer
(D) \(250 \mathrm{~mA}\)
Step-by-step Solution
Detailed explanation
For a lossless transformer, with primary coil \(\left(I_P, V_P\right)\) and secondary coil \(\left(I_S, V_S\right)\), there has to be conservation of energy per unit time:
\(\begin{aligned} & I_P V_P=I_S V_S \\ & \Rightarrow \frac{I_P}{I_S}=\frac{V_S}{V_P} \\ & \therefore \frac{I_P}{2}=\frac{25}{500} \\ & \therefore I_P=\frac{1}{4} \mathrm{~A}=250 \mathrm{~mA}\end{aligned}\)
\(\begin{aligned} & I_P V_P=I_S V_S \\ & \Rightarrow \frac{I_P}{I_S}=\frac{V_S}{V_P} \\ & \therefore \frac{I_P}{2}=\frac{25}{500} \\ & \therefore I_P=\frac{1}{4} \mathrm{~A}=250 \mathrm{~mA}\end{aligned}\)
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