Math, asked by sandhysingh81, 4 months ago

a 130 ohm resistor and 40uf capacitor are connected in series to an AC source of angular frequency w the value of w for which the combination acts as a pure resistive load is​

Answers

Answered by amitnrw
4

Given :  a 130 ohm resistor and 40uf capacitor are connected in series to an AC source of angular frequency ω

To Find :  the value of ω for which the combination acts as a pure resistive load is​

Solution:

pure resistive load  if  

XL =  Xc  

=> 2πfL =  1/2πfC

Here only capacitor is given  so need to have infinite frequency which is not possible

We need inductor also in the circuit

Then 2πfL =  1/2πfC

ω = 2πf

=> ωL = 1/ωC

=> ω² = 1/LC

⇒ ω = 1/√LC

To Find the value of ω for which the combination acts as a pure resistive load   , we need inductor also in circuit.

Then ω = 1/√LC  can be calculated

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Answered by nirman95
4

Given:

A 130 Ohm resistor and 40 \mu F capacitor are connected in series to an AC source of angular frequency \omega.

To find:

Value of frequency for which circuit is purely resistive?

Calculation:

Now, in the alternating circuit, only resistance and capacitor has been provided (but inductor has not been given).

  • So, for the circuit to be purely resistive: The CAPACITIVE REACTANCE has to be zero.

 \therefore \: X_{C} = 0

 \implies \: \dfrac{1}{ \omega C} = 0

 \implies \: \omega = \dfrac{1}{C \times 0}

 \implies \: \omega = \infty \: hz

So, the frequency of the AC circuit has to be INFINITY (theoritically).

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