Question 29: AC fundamentals

Question

A transformer has 400 turns in its primary winding and 800 turns in its secondary winding. The primary voltage is $$20 \mathrm{~V}$$ and the load in the secondary is 4 ohm. The current in the primary, assuming it to be an ideal transformer, is

Accepted Answer

To solve this problem, we can use the principles of an ideal transformer along with OhmÃ¢ÂÂs law. HereÃ¢ÂÂs the step-by-step process:
Determine the Voltage Ratio:
An ideal transformer follows the relationship:
$$\frac{V_s}{V_p} = \frac{N_s}{N_p}$$
Given:
Number of primary turns, $$N_p = 400$$
Number of secondary turns, $$N_s = 800$$
Primary voltage, $$V_p = 20\, 	ext{V}$$
Compute the secondary voltage:
$$V_s = V_p 	imes \frac{N_s}{N_p} = 20\, 	ext{V} 	imes \frac{800}{400} = 20\, 	ext{V} 	imes 2 = 40\, 	ext{V}$$
Calculate the Secondary Current:
With the load resistance on the secondary being 4 ohms, use Ohm's law:
$$I_s = \frac{V_s}{R} = \frac{40\, 	ext{V}}{4\, \Omega} = 10\, 	ext{A}$$
Find the Primary Current Using Power Conservation:
For an ideal transformer, power is conserved (ignoring losses):
$$V_p 	imes I_p = V_s 	imes I_s$$
Solve for the primary current ($$I_p$$):
$$I_p = \frac{V_s 	imes I_s}{V_p} = \frac{40\, 	ext{V} 	imes 10\, 	ext{A}}{20\, 	ext{V}} = \frac{400}{20} = 20\, 	ext{A}$$
Thus, the primary current is 20 A, which corresponds to Option B.

Answer

40 A

Answer

20 A

Answer

4 A

Answer

2 A

AC fundamentals

A transformer has 400 turns in its primary winding and 800 turns in its secondary winding. The primary voltage is $20 \mathrm{~V}$ and the load in the secondary is 4 ohm. The current in the primary, assuming it to be an ideal transformer, is

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