Electro Magnetic Induction And Alternating Currents Question 187
Question: A capacitor of $ 10\mu F $ and an inductor of 1 H are joined in series. An ac of 50 Hz is applied to this combination. What is the impedance of the combination?
Options:
A) $ \frac{5({{\pi }^{2}}-5)}{\pi }\Omega $
B) $ \frac{10^{2}(10-{{\pi }^{2}})}{\pi }\Omega $
C) $ \frac{10({{\pi }^{2}}-5)}{\pi }\Omega $
D) $ \frac{5^{2}(10-{{\pi }^{2}})}{\pi }\Omega $
Show Answer
Answer:
Correct Answer: B
Solution:
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Here, $ X_{L}={\omega_{L}}=2\pi fL=2\pi \times 50\times 1=100\pi \Omega $
$ X_{c}=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{1}{2\pi \times 50\times 10\times {{10}^{-6}}}=\frac{10^{3}}{\pi }\Omega $
So, $ =| X_{L}-. X_{C} | .=| 100\pi -. \frac{10^{3}}{\pi } | .=| 10^{2}. [ \frac{{{\pi }^{2}}-10}{\pi } ] | .\Omega $
BETA
