Question

An air-cored transformer is assumed to be 100% efficient. The ratio of the secondary turns to the primary turns is 1:20. A 240V ac supply is connected to the primary coil and a 6Ω load is connected to the secondary coil. what is the current in the primary coil?

a. 0.10A
b. 0.14A
c. 2.0A
d. 40.0A

Answer 1

An air-cored transformer is assumed to be 100% efficient. The ratio of the secondary turns to the primary turns is 1:20. A 240V ac supply is connected to the primary coil and a 6Ω load is connected to the secondary coil. what is the current in the primary coil?

a. 0.10A

b. 0.14A

c. 2.0A

d. 40.0A

Solution

We are given:

$$\frac {N _ {S}}{N _ {P}} = \frac {1}{2 0}$$ $$V _ {P} = 2 4 0 \mathrm {V}$$ $$R _ {S} = 6 \Omega$$

For ideal transformer:

$$\frac {V _ {P}}{V _ {S}} = \frac {I _ {S}}{I _ {P}} = \frac {N _ {P}}{N _ {S}}$$

Voltage on secondary coil :

$$V _ {S} = \frac {V _ {P} N _ {S}}{N _ {P}}$$

Using Ohm's law for secondary coil :

$$I _ {S} = \frac {V _ {S}}{R _ {S}} = \frac {V _ {P} N _ {S}}{N _ {P} R _ {S}}$$

Using expression for ideal transformer get current in the primary coil:

$$I _ {P} = \frac {I _ {S} N _ {S}}{N _ {P}} = \frac {V _ {P} N _ {S}}{N _ {P} R _ {S}} \frac {N _ {S}}{N _ {P}} = \frac {V _ {P}}{R _ {S}} * \left(\frac {N _ {S}}{N _ {P}}\right) ^ {2}$$

Calculation:

$$I _ {P} = \frac {2 4 0}{6} * \left(\frac {1}{2 0}\right) ^ {2} = \mathbf {0 . 1 0 A}$$

Answer: a, 0.10A