Question 290: AC fundamentals

Question

An ideal transformer converts 220 a.c. to 3.3 kV a.c. to transmit a power of 4.4 kW . If primary coil has 600 turns then alternating current in secondary coil is

Accepted Answer

Given data:
Primary voltage, $V_p = 220~	ext{V}$
Secondary voltage, $V_s = 3.3~	ext{kV} = 3300~	ext{V}$
Power transmitted, $P = 4.4~	ext{kW} = 4400~	ext{W}$
Number of primary turns, $N_p = 600$
Number of secondary turns, $N_s =$ (not required for this question)
Current in secondary coil, $I_s =$ ?
Step 1: Power in an ideal transformer
For an ideal transformer, power in primary = power in secondary:
$ P_p = P_s $
$ V_p I_p = V_s I_s $
Step 2: Using the given power
Here, total power transmitted = $P = 4400~	ext{W}$
$ V_s I_s = 4400 $
Step 3: Substitute the secondary voltage
$ 3300 	imes I_s = 4400 $
Step 4: Calculate $I_s$
$ I_s = \frac{4400}{3300} $
$ I_s = \frac{44}{33} $
$ I_s = \frac{4}{3}~	ext{A} $
Step 5: Final answer
The alternating current in secondary coil is:
$ \boxed{\frac{4}{3}~	ext{A}} $
So, the correct option is
Option C
.

Answer

$\frac{5}{3} \mathrm{~A}$

Answer

$\frac{1}{4} \mathrm{~A}$

Answer

$\frac{4}{3} \mathrm{~A}$

Answer

$\frac{2}{3} \mathrm{~A}$

AC fundamentals

An ideal transformer converts 220 a.c. to 3.3 kV a.c. to transmit a power of 4.4 kW . If primary coil has 600 turns then alternating current in secondary coil is

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