A source of e.m.f 110v and frequency 60hz is connected to a resistor, an indicator and a capacitor in series. When the current in the capacitor is 2A.the potential difference across the resistor is 80v and that across the inductor is 40v.calculate the -(1)potential difference across the capacitor (2)capacitance of the capacitor 3 inductance of the inductor
The current is the same in all three since they are in series, 2 amps.
Say i = 2 cos w t
where w = 2 pi f = 120 pi
then
volts across resistor = iR = 2R cos wt
We measure 80 volts, no phase given so assume 2R=80 and R = 40
volts across inductor = Ldi/dt = -240piL sin wt
again no phase given so assume 240piL = 40 and L = .053 H
volts across capacitor = q/C = integral i dt /C = [1/(60piC)] sin wt
I like your lesson
2) 1\2*3.142*60*115.5*10^-6
= 46uF
To find the potential difference across the capacitor, the capacitance of the capacitor, and the inductance of the inductor, we can use the given information and apply basic circuit analysis principles.
Let's start with the first question:
1. Potential difference across the capacitor:
The potential difference across the capacitor can be calculated using Ohm's Law. In a series circuit, the total voltage is equal to the sum of the individual voltage drops across each component.
Given:
EMF (E) = 110V
Voltage across the resistor (Vr) = 80V
Voltage across the inductor (Vi) = 40V
The potential difference across the capacitor (Vc) can be calculated as:
Vc = E - Vr - Vi
= 110V - 80V - 40V
= 30V
So, the potential difference across the capacitor is 30V.
Moving on to the second question:
2. Capacitance of the capacitor:
The current flowing through a capacitor (Ic) is given by the formula Ic = C * dV/dt, where C is the capacitance and dV/dt is the rate of change of voltage.
Given:
Current in the capacitor (Ic) = 2A
Frequency (f) = 60Hz
Since the current flowing through the capacitor is given, we can rearrange the formula to solve for capacitance:
C = Ic / (f * dV/dt)
Since the rate of change of voltage (dV/dt) is not provided, we cannot determine the capacitance without additional information.
Finally, let's solve the third question:
3. Inductance of the inductor:
The potential difference across an inductor (VL) can be calculated using Ohm's Law.
Given:
Voltage across the inductor (Vi) = 40V
Current in the inductor (I) = 2A
Using Ohm's Law, we can write:
VL = I * L
Rearranging the formula to solve for inductance (L):
L = VL / I
= 40V / 2A
= 20Ω
So, the inductance of the inductor is 20Ω.
To summarize:
1. The potential difference across the capacitor is 30V.
2. The capacitance of the capacitor cannot be determined without additional information.
3. The inductance of the inductor is 20Ω.
Now we measure 110 volts across the whole circuit and that must be the sum of the three voltages.
Therefore
110 sin(wt+phi) = 80 cos wt - 40 sin wt + [1/(60 pi C)] sin wt
110 sin(wt+phi)= 80 cos wt + [-40+ 1/(60piC)]sin wt
110[sinwt cosphi + coswt sinphi] =80 cos wt + [-40+ 1/(60piC)]sin wt
sin wt terms:
110 cos phi = [-40+ 1/(60piC)]
cos wt terms:
110 sin phi = 80 solver for phi, then use in above to solve for C