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 $



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