Electronics - Capacitors - Discussion
Discussion Forum : Capacitors - General Questions (Q.No. 59)
59.
What is the total current for a 5 MHz, 1 Vac circuit that consists of a 27 pF capacitor and a 1 k
resistor connected in parallel?
resistor connected in parallel?459 
A
A647 A
A1.31 mA
1.85 mA
Discussion:
7 comments Page 1 of 1.
Ansh said:
10 years ago
Can anyone explain what is it = ^?
(1)
Sumit aole said:
1 decade ago
Here the total current formula is I = sqrt(Ic^2 +Ir^2);
I= sqrt(v/Xc^2+v/r^2);
i = sqrt (10^-6 + 1/8.4823*10^-4);
i=sqrt(1178.92+10^-6);
i=1.31129*10^-3;
I= sqrt(v/Xc^2+v/r^2);
i = sqrt (10^-6 + 1/8.4823*10^-4);
i=sqrt(1178.92+10^-6);
i=1.31129*10^-3;
Alekhya said:
10 years ago
It means power like 10 power 6.
Leelavathi said:
10 years ago
Can anyone please elaborate this problem?
Alizwel said:
9 years ago
Get the total impedance of the parallel circuit then use ohms law.
Vishnu said:
8 years ago
I can't get it. Can anyone elaborate?
Gabbie said:
6 years ago
ITotal = sqrt [ (V/R)^2 + (V/Xc)^2 ].
ITotal = sqrt [ (1/1000)^2 + (2pi(5MHz)(27pico))^2].
NOTE:
XC= 1 over 1/2piFC.
Or in simplest term = 2piFC.
So it becomes ITotal = Sqrt [ V/R)^2 + (2piFC)^2].
ITotal = 1.31mA.
ITotal = sqrt [ (1/1000)^2 + (2pi(5MHz)(27pico))^2].
NOTE:
XC= 1 over 1/2piFC.
Or in simplest term = 2piFC.
So it becomes ITotal = Sqrt [ V/R)^2 + (2piFC)^2].
ITotal = 1.31mA.
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