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An ideal transformer converts 220 V AC t...

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

A

`1/3A`

B

`4/3A`

C

`5/3A`

D

`7/3A`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will use the properties of an ideal transformer and the formulas associated with it. ### Step 1: Identify the given values - Primary Voltage (Vp) = 220 V - Secondary Voltage (Vs) = 3.3 kV = 3300 V (since 1 kV = 1000 V) - Power (P) = 4.4 kW = 4400 W (since 1 kW = 1000 W) - Number of turns in the primary coil (Np) = 600 turns ### Step 2: Use the power formula to find the primary current (Ip) The power in an ideal transformer is given by the formula: \[ P = Vp \times Ip \] From this, we can rearrange to find the primary current (Ip): \[ Ip = \frac{P}{Vp} \] Substituting the known values: \[ Ip = \frac{4400 \, \text{W}}{220 \, \text{V}} \] \[ Ip = 20 \, \text{A} \] ### Step 3: Use the transformer ratio to find the secondary current (Is) For an ideal transformer, the relationship between voltages and currents is given by: \[ \frac{Vs}{Vp} = \frac{Ip}{Is} \] From this, we can express the secondary current (Is) as: \[ Is = \frac{Vp}{Vs} \times Ip \] Substituting the known values: \[ Is = \frac{220 \, \text{V}}{3300 \, \text{V}} \times 20 \, \text{A} \] \[ Is = \frac{220}{3300} \times 20 \] \[ Is = \frac{220 \times 20}{3300} \] ### Step 4: Simplify the expression Now, simplify the fraction: \[ Is = \frac{4400}{3300} \] Dividing both the numerator and the denominator by 1100 gives: \[ Is = \frac{4}{3} \, \text{A} \] ### Final Answer The alternating current in the secondary coil (Is) is: \[ Is = \frac{4}{3} \, \text{A} \]
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