Electronics - RLC Circuits and Resonance - Discussion
Discussion Forum : RLC Circuits and Resonance - General Questions (Q.No. 21)
21.
What is the true power consumed in a 30 V series RLC circuit if Z = 20 ohms and R = 10 ohms?
Discussion:
14 comments Page 1 of 2.
Afash said:
1 decade ago
Friends Remember one thing. When the value of Z given in RLC ctk it means that the total R of the circuit is Z. as like:
Z= sqrt of R^2 + (Xl -Xc)^2.
So there is no need to find Resistance of the ckt.
Just like:
I = v/Z.
1.5 A.
Than T.Power = I^2 *R.
So you will get the answer.
Z= sqrt of R^2 + (Xl -Xc)^2.
So there is no need to find Resistance of the ckt.
Just like:
I = v/Z.
1.5 A.
Than T.Power = I^2 *R.
So you will get the answer.
Anjay pal said:
1 decade ago
@Neeraja your calculation is wrong.
Zt=sqrt(Z2+R2)
Zt=sqrt{(20)2+(10)2}, =sqrt(400+100),=sqrt(500),
zt=22.36ohms
It=V/Zt
It=30/22.36
It=1.34amps
true power=It2*R
(1.34)2*10
1.7956*10
17.95W
But this wrong answer.
Zt=sqrt(Z2+R2)
Zt=sqrt{(20)2+(10)2}, =sqrt(400+100),=sqrt(500),
zt=22.36ohms
It=V/Zt
It=30/22.36
It=1.34amps
true power=It2*R
(1.34)2*10
1.7956*10
17.95W
But this wrong answer.
Neeraja said:
1 decade ago
Zt=sqrt(z2+R2) where Zt equals to around 20.2
Zt = 20.2ohms
V= 30v
It=V/Zt= 1.5amps
R = 10ohms
true Power = It2*R
(1.5)2* 10
2.25*10
22.5W
Zt = 20.2ohms
V= 30v
It=V/Zt= 1.5amps
R = 10ohms
true Power = It2*R
(1.5)2* 10
2.25*10
22.5W
Hriday said:
9 years ago
@Neeraja.
z itself includes r so, do not solve to zt.
z is already given and the z value is used to calculate the current whereas r is used to calculate the power.
z itself includes r so, do not solve to zt.
z is already given and the z value is used to calculate the current whereas r is used to calculate the power.
Akshay said:
1 decade ago
Circuit current = v/z = 30/20 = 3/2.
True power = vicos (theta).
Where cos (theta) = r/z = 10/20 = 5/10.
True power = 30*3/2*5/10 = 22.5w.
True power = vicos (theta).
Where cos (theta) = r/z = 10/20 = 5/10.
True power = 30*3/2*5/10 = 22.5w.
Simi said:
9 years ago
True power= I^2 * R.
Reactive Power = I^2 * X.
Apparent Power = I^2 * Z.
Here, I = V/Z = 30 / 20 = 3/2.
True power =(3/2)^2 * 10 = 22.5W.
Reactive Power = I^2 * X.
Apparent Power = I^2 * Z.
Here, I = V/Z = 30 / 20 = 3/2.
True power =(3/2)^2 * 10 = 22.5W.
(1)
Marvin said:
4 years ago
G = Re(Z) = R/Z^2 = 0.025 Ohm;
Active power dissipates only on active resistance (conductivity)
P = U^2G = 30^2 * 0.025 = 22.5 W.
Active power dissipates only on active resistance (conductivity)
P = U^2G = 30^2 * 0.025 = 22.5 W.
(1)
Ram mohan said:
1 decade ago
Because circuit impedence =20 ohm
So current in circuit = v/z = 30/20 = 1.5 amp
True power = I^2XR = (1.5)^2 X 10 = 22.5 Watt
So current in circuit = v/z = 30/20 = 1.5 amp
True power = I^2XR = (1.5)^2 X 10 = 22.5 Watt
Dhaval pujara said:
8 years ago
Here given,
V = 30v
Z = 20ohms
I = v/z
I=??
Then,
True power(p) = i^2*r.
V = 30v
Z = 20ohms
I = v/z
I=??
Then,
True power(p) = i^2*r.
Amritpal singh said:
1 decade ago
Then why the answer is not coming by p = I2R = VI = 30*1.5 = 45.
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