Question:medium

In a step-up transformer the voltage in the primary is $220\mathrm{V}$ and the current is $5\mathrm{A}$. The secondary voltage is found to be $22000\mathrm{V}$. The current in the secondary (neglect losses) is

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In a transformer, $V_p I_p = V_s I_s$ (neglecting losses).
Updated On: May 3, 2026
  • $5\mathrm{A}$
  • $50\mathrm{A}$
  • $500\mathrm{A}$
  • $0.05\mathrm{A}$
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The Correct Option is D

Solution and Explanation

To solve the problem regarding the step-up transformer, we use the concept of conservation of power. In an ideal transformer (neglecting losses), the power input to the primary coil equals the power output from the secondary coil. The formula connecting the voltage and current in the primary and secondary coils is given by:

\(P_{\text{primary}} = P_{\text{secondary}}\)

Here, power (\(P\)) is the product of voltage (\(V\)) and current (\(I\)):

\(V_{\text{primary}} \times I_{\text{primary}} = V_{\text{secondary}} \times I_{\text{secondary}}\)

Given:

  • \(V_{\text{primary}} = 220\,\mathrm{V}\)
  • \(I_{\text{primary}} = 5\,\mathrm{A}\)
  • \(V_{\text{secondary}} = 22000\,\mathrm{V}\)

We need to find \(I_{\text{secondary}}\).

Substitute the given values into the equation:

\(220\,\mathrm{V} \times 5\,\mathrm{A} = 22000\,\mathrm{V} \times I_{\text{secondary}}\)

Calculate the power in the primary coil:

\(P_{\text{primary}} = 220 \times 5 = 1100\,\mathrm{W}\)

Set this equal to the power in the secondary coil and solve for \(I_{\text{secondary}}\):

\(1100\,\mathrm{W} = 22000\,\mathrm{V} \times I_{\text{secondary}}\)

\(I_{\text{secondary}} = \frac{1100}{22000} = 0.05\,\mathrm{A}\)

Therefore, the current in the secondary coil is \(0.05\,\mathrm{A}\).

The correct answer is

\(0.05\,\mathrm{A}\) 

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