Question 32: AC fundamentals

Question

A transformer which steps down $$330 \mathrm{~V}$$ to $$33 \mathrm{~V}$$ is to operate a device having impedance $$110 \Omega$$. The current drawn by the primary coil of the transformer is :

Accepted Answer

First, let's identify the voltage transformation ratio for the transformer. The primary voltage $$V_p$$ is $$330 \mathrm{~V}$$ and the secondary voltage $$V_s$$ is $$33 \mathrm{~V}$$. The voltage transformation ratio $$k$$ is given by:
$$ k = \frac{V_p}{V_s} = \frac{330}{33} = 10 $$
This means the primary voltage is 10 times the secondary voltage.
Next, we need to determine the current in the secondary coil, which can be found using Ohm's Law. The impedance of the device connected to the secondary coil is given as $$110 \Omega$$, and the voltage across it is $$33 \mathrm{~V}$$:
$$ I_s = \frac{V_s}{Z} = \frac{33}{110} = 0.3 \mathrm{~A} $$
Now, using the current transformation relation in a transformer, which is inversely proportional to the voltage transformation ratio:
$$ I_p = \frac{I_s}{k} = \frac{0.3}{10} = 0.03 \mathrm{~A} $$
The current drawn by the primary coil of the transformer is therefore $$0.03 \mathrm{~A}$$.
So, the correct answer is:
Option B:
0.03 A

Answer

0.3 A

Answer

0.03 A

Answer

3 A

Answer

1.5 A

AC fundamentals

A transformer which steps down $330 \mathrm{~V} to 33 \mathrm{~V} is to operate a device having impedance 110 \Omega$. The current drawn by the primary coil of the transformer is :

Choose the correct answer: