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:
Discussion:
37 comments Page 1 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.
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.
Chandrasekhar said:
1 decade ago
Let us for suppose that an i current is flowed in to the circuit such that "the voltage developed across the series pair of resistance and capacitor is == I(R+jXc)
The voltage = IR + j(IXc) = voltage acc resistance + j(voltage acrosscapacitor)
= 20+j40
= sqrt(20*20 + 40 *40)
= 44.72
The voltage = IR + j(IXc) = voltage acc resistance + j(voltage acrosscapacitor)
= 20+j40
= sqrt(20*20 + 40 *40)
= 44.72
Saurabh kumar said:
1 decade ago
Let us for suppose that an i current is flowed in to the circuit such that "the voltage developed across the series pair of resistance and capacitor is == I(R+jXc)
The voltage = IR + j(IXc) = voltage acc resistance + j(voltage acrosscapacitor)
= 20+j40
= sqrt(20*20 + 40 *40)
= 44.72
The voltage = IR + j(IXc) = voltage acc resistance + j(voltage acrosscapacitor)
= 20+j40
= sqrt(20*20 + 40 *40)
= 44.72
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.
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.
Gabbie said:
6 years ago
For series RC circuit, the formula is E= √(E resistive component^2 + E capacitive component^2)
E= √(20^2 + 40^2) = 44.72 V approx. 45 V.
The formula is from power triangle.
E= √(20^2 + 40^2) = 44.72 V approx. 45 V.
The formula is from power triangle.
(4)
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.
So V = IR+jIX which in vector form so finally magnitude form is V = square root of V, VC.
Izhar Rao said:
1 decade ago
Vt=(Vr^2+Vc^2)^1/2
Vt=(20^2+40^2)^1/2
Vt=10(2^2+4^2)^1/2
Vt=10(4+16)^1/2
Vt=10(20)^1/2
Vt=10x4.472
Vt=44.72
Vt=45 volt.
Vt=(20^2+40^2)^1/2
Vt=10(2^2+4^2)^1/2
Vt=10(4+16)^1/2
Vt=10(20)^1/2
Vt=10x4.472
Vt=44.72
Vt=45 volt.
Jayendra said:
1 decade ago
Here, rc series circuit therefore as formula vt= squareroot of vr*vr+vc*vc=400.
1600=44. 72=45v.
Thats the answer.
1600=44. 72=45v.
Thats the answer.
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.
Here above solution is two voltages are taken into count.
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.
To find the magnitude of the above expression:
Val = sqrt(20*20 + 40*40) = 44.7213 == 45.6.
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