A coil of 50 ohm resistance and of 150mH inductance is connected in parallel with a capacitor 50microF. The source voltage is 100 sin(wt+30). What is the equation line current.
Pa help solutions
Ohm my goodness, it seems like we have an electrifying problem on our hands! Let's juggle through the calculations and try to find the equation for the line current.
In a parallel circuit, the line current is equal to the sum of the currents flowing through the individual components. Since the coil and capacitor are connected in parallel, we need to calculate the current across each of them separately.
For the coil, we know its resistance (R = 50 ohms) and inductance (L = 150 mH = 0.15 H). The voltage across the coil is the same as the source voltage, so we can use Ohm's law and the formula for the current in an inductive circuit to find the coil current.
Using Ohm's law: V = IR
We have 100 sin(wt + 30) = I_coil * 50
Therefore, I_coil = (100 sin(wt + 30)) / 50
Now, for the capacitor, we can use the formula for the current in a capacitive circuit, which is the derivative of the voltage with respect to time, multiplied by the capacitance.
Using the formula: I_capacitor = C * dV/dt
We have I_capacitor = 50 μF * d/dt (100 sin(wt + 30))
Now, all that's left is to add the two currents together for the total line current:
I_line = I_coil + I_capacitor
And there you have it, the equation for the line current! I hope I didn't shock you with all these calculations. Remember, humor is my current specialty!
To find the equation for line current, we can start by finding the impedance of the parallel combination of the coil and capacitor.
The impedance of the coil (ZL) is given by:
ZL = jwL
Where:
j is the imaginary unit (√(-1))
w is the angular frequency (2πf)
L is the inductance
In this case, w = 2πf = 2π * (1/T)
T = 1/f, where f is the frequency
Given:
L = 150 mH = 150 × 10^(-3) H
f = 1/T
T = 1/100 = 0.01 s
So, w = 2π * (1/0.01) = 628.32 rad/s
Substituting the values, we get:
ZL = j * 628.32 * 150 × 10^(-3)
ZL ≈ j * 94.25
The impedance of the capacitor (ZC) is given by:
ZC = 1/(jwC)
Where:
C is the capacitance
Given:
C = 50 μF = 50 × 10^(-6) F
Substituting the values, we get:
ZC = 1/(j * 628.32 * 50 × 10^(-6))
ZC ≈ 1/(j * 0.031416)
To find the total impedance (Z), we can use the formula for the parallel combination of two impedances:
1/Z = 1/ZL + 1/ZC
Substituting the values, we get:
1/Z ≈ 1/(j * 94.25) + 1/(j * 0.031416)
To simplify the expression, we can take the reciprocals and then add them together:
1/Z ≈ j/(94.25 * (j * 0.031416)) + j/(j * 0.031416)
1/Z ≈ j/(2.9656 + 1)
Adding the fractions, we get:
1/Z ≈ j/(3.9656)
To find Z, we can take the reciprocal:
Z ≈ 3.9656/j
The total impedance is approximately 3.9656/j.
Now, to find the line current (IL), we can use Ohm's law:
IL = VS/Z
Given:
VS = 100 sin(wt+30)
Substituting the values, we get:
IL = (100 sin(wt+30))/(3.9656/j)
Simplifying the expression, we multiply the expression by the complex conjugate of the denominator:
IL = (100 sin(wt+30))/(3.9656/j) * (j/(-j))
IL = -100j sin(wt+30)/3.9656
Therefore, the equation for the line current is approximately -100j sin(wt+30)/3.9656.
To determine the equation for the line current in this circuit, we need to analyze the behavior of the components and calculate the impedance of the circuit.
First, let's analyze the components in the circuit:
1. The resistance of the coil is given as 50 ohms.
2. The inductance of the coil is given as 150mH, which is equivalent to 0.15H.
3. The capacitance of the capacitor is given as 50 microfarads, which is equivalent to 50 × 10^(-6) farads.
4. The source voltage is given by V(t) = 100sin(wt + 30), where w represents the angular frequency in radians per second and t represents time.
Next, we need to calculate the impedance (Z) of the circuit by combining the impedance of the coil (ZL) and the impedance of the capacitor (ZC) in parallel.
The impedance of the capacitor (ZC) can be calculated using the formula ZC = 1 / (jwC), where j is the imaginary unit.
Substituting the given values, we get ZC = 1 / (j * w * 50 × 10^(-6)).
The impedance of the coil (ZL) can be calculated using the formula ZL = jwL, where j is the imaginary unit.
Substituting the given values, we get ZL = j * w * 0.15.
Now, we can calculate the net impedance (Z) of the circuit by combining the impedance of the coil and the impedance of the capacitor in parallel.
Using the formula for the inverse of the sum of the inverses: 1 / Z = 1 / ZL + 1 / ZC,
we get 1 / Z = 1 / (j * w * 0.15) + 1 / (j * w * 50 × 10^(-6)).
To simplify the expression, we need to find a common denominator by multiplying the first term by (jw * 50 × 10^(-6)) and the second term by (jw * 0.15).
After simplification, we get 1 / Z = (0.15 − j * w * 50 × 10^(-6)) / (j * w * 0.15 * 50 × 10^(-6)).
To calculate the inverse of Z, we take the reciprocal of the expression: Z = (j * w * 0.15 * 50 × 10^(-6)) / (0.15 − j * w * 50 × 10^(-6)).
Finally, to find the line current (I), we divide the source voltage (V) by the impedance (Z):
I = V / Z = (100sin(wt + 30)) / ((j * w * 0.15 * 50 × 10^(-6)) / (0.15 − j * w * 50 × 10^(-6))).
This equation represents the line current as a function of time (t).
Please note that the final equation is simplified using basic algebraic operations and the formulas for impedance in inductive and capacitive components.