Electronics - Capacitors - Discussion

Discussion Forum : Capacitors - General Questions (Q.No. 1)
1.
In a 20 Vac series RC circuit, if 20 V is measured across the resistor and 40 V is measured across the capacitor, the applied voltage is:
45 V
50 V
55 V
60 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
37 comments Page 2 of 4.

Edmond Kong said:   9 years ago
The equation is V = V(R) - jV(C).

Voltage across the resistor, V(R), is on the positive X-axis.
Voltage across the capacitor, V(C), is on the negative Y-axis (because of the -j).
To find the resultant voltage, use the Pythagorean Theorem (Right Triangle Formula):

Z^2 = X^2 + Y^2
Z = sqrt(X^2 + Y^2), where X = 20V and Y = 40V

Therefore, in theory, Z = 44.72136 V.

Anil said:   10 years ago
What about 20 Vac applied (in equation)?

Shantanu behera said:   1 decade ago
Formula is total impedance = square root of (xr2+xl2).

Here above solution is two voltages are taken into count.

Anish kumar said:   1 decade ago
Voltage developed across resistor = 20 v.

Voltage developed across capacitor = 40 v.

Effective voltage = 20 + 40j.

Magnitude of effective voltage = sqrt(20+40j) = 44.72 v == 45 v.

Effective voltage = Input voltage supplied.

Valli said:   1 decade ago
I didn't understand the solution can anybody solve little more simple.

Ravi said:   1 decade ago
In A.C circuits Z = R+jX. So V = IZ.

So V = IR+jIX which in vector form so finally magnitude form is V = square root of V, VC.

Love said:   1 decade ago
But the impedance of capacitor is counted as 1/jwc.

Vinayak birje said:   1 decade ago
20Vac is for what?

Brijeshyadav said:   1 decade ago
In the starting of question 20Vac circuit accounts for what?

Shiva said:   1 decade ago
(20 + j40).

To find the magnitude of the above expression:

Val = sqrt(20*20 + 40*40) = 44.7213 == 45.6.


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