Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 70)
70.
What is the impedance of an ac RC parallel circuit if the resistance is 12 ohms and the capacitive reactance equals 5 ohms?
Discussion:
24 comments Page 2 of 3.
Dsce said:
9 years ago
Z = R + jXc
= 12 + j5
= sqrt(12^2 + 5^2)
= sqrt144 + 25
= 13 ohm.
= 12 + j5
= sqrt(12^2 + 5^2)
= sqrt144 + 25
= 13 ohm.
Sagar dukre said:
9 years ago
I agree with you @Dsce.
The answer is 13ohm.
The answer is 13ohm.
Samir said:
9 years ago
It's parallel circuit connection that's why the equation for it,
Z = RXC / SQRT OF (R^2 + XC^2).
And for series connection as you calculated above method:
z = sqrt of r^2 + xc^2.
Thanks.
Z = RXC / SQRT OF (R^2 + XC^2).
And for series connection as you calculated above method:
z = sqrt of r^2 + xc^2.
Thanks.
BANAR said:
9 years ago
How it has 4.16? The correct answer is 13 ohm.
Apeksha said:
9 years ago
It's 4.615.
RXC/√xc^2+r^2.
You will get 4.615.
RXC/√xc^2+r^2.
You will get 4.615.
Riyu said:
8 years ago
I also agree that the answer is 13 ohm.
Prp said:
8 years ago
Impedance Z in parallel RC circuit.
1/Z= √[(1/R^2)+(1/Xc)^2].
=°[(1/144)+(1/25)]=0 .21656.
Z=1/0.21656= 4.617ohm.
1/Z= √[(1/R^2)+(1/Xc)^2].
=°[(1/144)+(1/25)]=0 .21656.
Z=1/0.21656= 4.617ohm.
Kate said:
8 years ago
Impedance of parallel RC circuit is:
Z = R*Xc/(sqrt(R^2+Xc^2).
Plug in all values then you'll get 4.615384615 ~ 4.61
Z = R*Xc/(sqrt(R^2+Xc^2).
Plug in all values then you'll get 4.615384615 ~ 4.61
Akshar said:
8 years ago
Impedance Z in parallel RC circuit.
1/Z = √[(1/R^2)+(1/Xc)^2].
=√[(1/144)+(1/25)]=0 .21656.
Z = 1/0.21656= 4.617ohm.
1/Z = √[(1/R^2)+(1/Xc)^2].
=√[(1/144)+(1/25)]=0 .21656.
Z = 1/0.21656= 4.617ohm.
Kaveri said:
8 years ago
The correct answer is 13 ohm.
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