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In a transformer the number of primary t...

In a transformer the number of primary turns is four times that of the secondary turns. Its primary is connected to an AC source of voltage V. Then

A

current through its secondary is about four times that of the current through its primary.

B

voltage across its secondary is about four times that of the voltage across its primary.

C

voltage across its secondary is about two times that of the voltage across the primary.

D

voltage across its secondary is about `(1)/(2sqrt2)` times of the voltage across its primary.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the relationship between the primary and secondary sides of a transformer. Here’s a step-by-step solution: ### Step 1: Understand the Transformer Relationships A transformer operates on the principle of electromagnetic induction and has two main relationships: 1. The voltage ratio between the primary (Vp) and secondary (Vs) is proportional to the turns ratio (N): \[ \frac{V_p}{V_s} = \frac{N_p}{N_s} \] 2. The power input to the primary (Pp) is equal to the power output from the secondary (Ps): \[ P_p = P_s \implies V_p \cdot I_p = V_s \cdot I_s \] ### Step 2: Given Information - The number of primary turns \(N_p\) is four times the number of secondary turns \(N_s\): \[ N_p = 4N_s \] - The primary voltage \(V_p\) is connected to an AC source of voltage \(V\). ### Step 3: Find the Voltage Relationship Using the turns ratio: \[ \frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{4N_s}{N_s} = 4 \] Thus, we can express the secondary voltage in terms of the primary voltage: \[ V_s = \frac{V_p}{4} \] ### Step 4: Find the Current Relationship Using the power relationship: \[ V_p \cdot I_p = V_s \cdot I_s \] Substituting \(V_s\) from the previous step: \[ V_p \cdot I_p = \left(\frac{V_p}{4}\right) \cdot I_s \] Now, we can cancel \(V_p\) (assuming \(V_p \neq 0\)): \[ I_p = \frac{I_s}{4} \] Rearranging gives: \[ I_s = 4I_p \] ### Step 5: Conclusion The current in the secondary \(I_s\) is four times the current in the primary \(I_p\): \[ I_s = 4I_p \] Thus, the correct answer is that the current in the secondary is about four times that of the current through the primary. ### Final Answer The current in the secondary is four times that of the current in the primary. ---
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