Electro Magnetic Induction And Alternating Currents Question 30
Question: In a series circuit R = 300 W, L = 0.9 H, C = 2.0 mF and w = 1000 rad/sec. The impedance of the circuit is [MP PMT 1995]
Options:
A) 1300 W
B) 900 W
C) 500 W
D) 400 W
Show Answer
Answer:
Correct Answer: C
Solution:
For series R-L-C circuit,
$ Z=\sqrt{R^{2}+{{(X_{L}-X_{C})}^{2}}} $
$ =\sqrt{{{(300)}^{2}}+{{( 1000\times 0.9-\frac{10^{6}}{1000\times 2} )}^{2}}}=500\Omega $
