Question

A step up transformer operates on $220\text{ V}$ and supplies current of $2\text{ A}$. The ratio of primary and secondary windings is $1 : 20$. The current in the primary is

• A. $5\text{ A}$
• B. $2\text{ A}$
• C. $40\text{ A}$
• D. $20\text{ A}$

Hint:

A step-up transformer increases voltage but must proportionally step down the current to satisfy conservation of energy. Since the secondary side has 20 times more loops, its voltage scales up by 20, which forces its current to drop by a factor of 20. Therefore, the primary current must simply be $2\text{ A} \times 20 = 40\text{ A}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
An ideal step-up transformer has an alternating input voltage fed to its primary coil ($V_p = 220\text{ V}$).
The secondary coil outputs a current ($I_s = 2\text{ A}$). The turns ratio of the primary to secondary winding is given as $\frac{N_p}{N_s} = \frac{1}{20}$. We need to compute the corresponding current running through the primary circuit ($I_p$).

Step 2: Key Formula or Approach:
For an ideal transformer with zero net power dissipation losses, the power input to the primary coil matches the power output from the secondary coil ($P_p = P_s$):
$$V_p I_p = V_s I_s \implies \frac{I_p}{I_s} = \frac{V_s}{V_p}$$ The voltage ratio scales proportionally with the coil winding turn numbers:
$$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$ Combining these gives the inverse relationship between coil turns and currents:
$$\frac{I_p}{I_s} = \frac{N_s}{N_p} \implies I_p = I_s \left(\frac{N_s}{N_p}\right)$$

Step 3: Detailed Explanation:
The given winding turns ratio is $\frac{N_p}{N_s} = \frac{1}{20}$.
Inverting this fraction for our current balance formula gives:
$$\frac{N_s}{N_p} = \frac{20}{1} = 20$$ Now substitute the output current ($I_s = 2\text{ A}$) and this turn ratio into our current expression:
$$I_p = 2\text{ A} \times 20 = 40\text{ A}$$

Step 4: Final Answer:
The current in the primary winding is $40\text{ A}$, which maps to option (C).