RCL Circuits electricity!?
A 50V, 50Hz supply is connected to a series circuit consisting of a resistance, a 20mH inductor and a capacitor.
a) Find the value of the capacitance and the power dissipated in the circuit when:
(i) The current has a magnitude of 25A and leads the voltage by 45degrees.
(ii) The current has a magnitude of 25A and lags the voltage by 45degress.
(iii) The current and voltage are in phase.
b) If the frequency of the supply can be increased from 50Hz, using the value of the capacitance found in (i), find the frequency other than 50Hz when the current will have a magnitude of 25A.
I wonder if the resonant frequency is 50hz
freson=1/2PI sqrtLC
solve for C
Power=mag (V*I) cosAnglebetween
a1. Z(Ohms) = 50/25A[45o] = 2[-45o] = 2*Cos(-45) + j2*sin(-45) = 1.414 - j1.414.
Xl+Xc = -1.414, 6.28*50*0.020+Xc = -1.414, Xc = -7.69 Ohms.
C = 1/W*Xc = 1/(314.2*7.69) = 4.14*10^-4 Farads = 414 uF.
P = I^2*R = 25^2 * 1.414 = 884 Watts.
a2. Same procedure as a1.
a3. Z(Ohms) = E/I = 50/25A[0o] = 2[0o]. Xl+Xc = 0, Xc = -Xl = -WL = -314.2 * 0.020 = -6.28 Ohms.
C = 1/W*Xc = 1/(314.2*6.28) = 5.1*10^-4 Farads = 510 uF.
P = I^2*R = 25^2 * 2 = 1250 Watts.
P = I^2*R = 25^2 * 1.414 = 884 Watts.
To solve this problem, we need to use the concepts of impedance, reactance, and power in RCL circuits. Let's break down the steps to find the values of capacitance and power dissipated in the circuit for different scenarios.
a) To find the value of capacitance and power when the current has a magnitude of 25A and leads the voltage by 45 degrees:
Step 1: Calculate the reactance of the inductor (XL) and the resistance (R):
- The reactance of the inductor is given by XL = 2πfL, where f is the frequency and L is the inductance.
- For our problem, f = 50Hz and L = 20mH = 0.02H. Substituting the values, we get XL = 2π(50)(0.02) = 6.28 ohms.
Step 2: Calculate the total impedance (Z) of the circuit:
- In a series circuit, the total impedance is the sum of resistive (R) and reactive (XL or XC) impedance. Since the current leads the voltage, we have to consider the impedance due to the capacitor:
- In a capacitor, the reactance (XC) is given by XC = 1/(2πfC), where C is the capacitance.
- We will assume the capacitance is C. Thus, XC = 1/(2π(50)C) = 1/(100πC).
- Since the circuit has a series combination of R, XL, and XC, the total impedance Z is given by Z = √(R^2 + (XL - XC)^2).
- Substituting the values, we get Z = √(R^2 + (6.28 - 1/(100πC))^2).
Step 3: Calculate the capacitance (C):
- We know that the magnitude of the current (I) is 25A, so to find the value of C, we need to find the frequency where the total impedance Z is 25 ohms.
- Substitute Z = 25 in the above equation and solve for C.
Step 4: Calculate the power dissipated in the circuit:
- The power (P) dissipated in the circuit is given by P = IV, where I is the magnitude of the current and V is the magnitude of the voltage at that power factor angle θ.
- Since the current leads the voltage, θ = 45 degrees. Hence, we will use P = I * V * cos(45).
Now let's go through the steps (i), (ii), and (iii):
(i) The current has a magnitude of 25A and leads the voltage by 45 degrees:
- Follow the steps mentioned above to calculate the capacitance (C) and the power dissipated in the circuit.
(ii) The current has a magnitude of 25A and lags the voltage by 45 degrees:
- Repeat the steps, but this time, consider the impedance of the capacitor (XC) and find the value of C where Z = 25 ohms.
(iii) The current and voltage are in phase:
- In this scenario, XC = XL, and therefore, we have to solve Z = 25 ohms by considering the impedance due to both inductor and capacitor.
b) To find the frequency other than 50Hz when the current will have a magnitude of 25A:
- Use the value of capacitance (C) found in part (i) and rearrange the equation for XC = 1/(2πfC) to solve for the frequency other than 50Hz. Substitute XC = 25 ohms and C to find the new frequency.
Remember to use appropriate units in your calculations and always double-check your equations and math.