Question

in an ac circuit 600 mH inductor and a 50 micro farad capacitor are connected in series with 10 ohm resistance​

Answer 1

Answer:

An Ac circuit with R=4h , WL = 3 the impedance is ?

Answer 2

Given:

Inductance (L) $= 600 mH = 600 * 10^{-3} mH$

Capacitance (C) $= 50 mF = 50*10^{-6} mF$

Resistance (R) = 10Ω

Frequency (f) = 50 Hz

Potential difference (V) = 230 V

To find:

Power dissipated in the circuit.

Solution :

Since, we have the Inductance, we can calculate reactance of the inductor.

Inductive reactance ($X_{L}$) = ωL

$X_{L}$ = 2 π f L

Substituting the values, we get

$X_{L} = 2 (3.14) 50 (600*10^{-3} ) \\X_{L} = 189 ohm$

Similarly,

Capacitive reactance ($X_{C}$) = 1/ωC

$X_{C} = \frac{1}{2* 3.14*50*50*10^{-6} }$

$X_{C} = 64 ohm$

Now,

Impedance (Z) = $\sqrt{(R)^{2}+(X_{L} -X_{C} )^{2} }$

Substituting the values, we get

$Z = \sqrt{(10)^{2} + (189-64)^{2} } \\Z = \sqrt{(100+(125)^{2} } \\Z= \sqrt{15725} \\Z= 125$

We have,

Current (I) in the circuit = $\frac{V}{Z}$

Therefore, $I = \frac{230}{125} \\I = 1.85 A$

Now,

Power dissipated (P) = $V_{rms}* I_{rms} * cos\alpha$

where, $cos\alpha =\frac{R}{Z}$

Since, we know Inductor and capacitor do not consume power. The power consumption is solemnly by resistor.

Substituting the values, we get

$P = \frac{230 * 1.85*10}{125} \\P = 34 W$

Final answer:

Hence, the power dissipated in the following LCR circuit will be 34W.

Although your question is incomplete, you might be referring to the question below.

An AC circuit containing 600 mH inductor and a 50 mF capacitor connected in series with a resistance 10Ω. They are connected to 230 V, 50 Hz AC supply. Obtain average power absorbed.