Step 1: Understanding the Question:
The question asks us to identify the correct statement regarding the operational parameters (turns, voltage, and current) of a step-up transformer.
Step 2: Key Formula or Approach:
For an ideal transformer, the transformation ratio is governed by the relation:
$$\frac{V_s}{V_p} = \frac{N_s}{N_p} = \frac{I_p}{I_s}$$
Where the subscripts $p$ and $s$ represent the primary and secondary coils respectively.
In a
step-up transformer, the secondary voltage is higher than the primary voltage ($V_s > V_p$), which means the secondary turn count must be greater than the primary turn count ($N_s > N_p$).
Step 3: Detailed Explanation:
By the conservation of energy, the total electrical power in an ideal transformer remains constant between the two stages ($P_p = P_s$):
$$V_p I_p = V_s I_s \implies \frac{I_p}{I_s} = \frac{V_s}{V_p}$$
Since a step-up transformer increases the output voltage ($V_s > V_p$), the ratio $\frac{V_s}{V_p} > 1$.
This requires that:
$$\frac{I_p}{I_s} > 1 \implies I_p > I_s$$
Therefore, the alternating current flowing in the primary circuit coil must be strictly greater than the current flowing in the secondary circuit coil. This maps perfectly to the description in option (C).
Step 4: Final Answer:
The correct statement is that the current in the primary coil is more than the current in the secondary coil, matching option (C).