Electro Magnetic Induction And Alternating Currents Question 209
Question: An ideal efficient transformer has a primary power input of 10 kW. The secondary current when the transformer is on load is 25 A. If the primary: secondary turns ratio is 8 : 1, then the potential difference applied to the primary coil is
Options:
A) $ \frac{10^{4}\times 8^{2}}{25}V $
B) $ \frac{10^{4}\times 8}{25}V $
C) $ \frac{10^{4}}{25\times 8}V $
D) $ \frac{10^{4}}{25\times 8^{2}}V $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ P=V_{1}i_{1}=V_{2}i_{2} $ $ V_{2}=\frac{10^{4}}{25} $ Now $ V_{1}=\frac{n_{1}}{n_{2}}\times V_{2}=\frac{8}{1}\times \frac{10^{4}}{25},V $
BETA
