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
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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Given:
A 130 Ohm resistor and 40 capacitor are connected in series to an AC source of angular frequency
.
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.
So, the frequency of the AC circuit has to be INFINITY (theoritically).