A resistor of 15 ohms and an inductor of 4H and capacitor of 25 micro F are connected in series across 230V AC supply calculate.
a.The frequency
b.current
c.voltage drop or pd across inductance.
I cannot calculate the freq. unless the reactance of the inductor or capacitor is given. I will assume a 230v.- 60 Hz supply.
a 60 Hz.
b. Xl = 2pi*F*L = 6.28*60*4 = 1508 Ohms.
Xc = 1/(2p*F*C) = 1/(6.28*60*25*10^-6=
106 Ohms.
Z = R + j(Xl-Xc)
Z = 15 + j(1508-106)
Z = 15 + j1402 = Rectangular form.
tanA = Xl/R = 1402/15 = 93.5
A = 89.4o
Z = Xl/sinA = 1402/sin89.4=1402.1 Ohms
@ 89.4o = Polar form.
I = V/Z = 230/1402.1=0.164A @ (-89.4o)
c. V = I * Xl = 0.164 * 1508 = 247.3 V.
To find the answers, follow these steps:
a. To calculate the frequency, use the formula:
frequency = 1 / (2 * π * square root (L * C))
where L is the inductance and C is the capacitance.
Plugging in the given values:
L = 4H
C = 25 × 10^(-6) F
frequency = 1 / (2 * π * square root (4 * 25 × 10^(-6)))
= 1 / (2 * 3.14 * square root (100 × 10^(-6)))
= 1 / (6.28 * square root(100 × 10^(-6)))
= 1 / (6.28 * 0.01)
= 1 / 0.0628
≈ 15.93 Hz
Therefore, the frequency is approximately 15.93 Hz.
b. To calculate the current, divide the voltage by the total impedance (Z) in the circuit. The impedance is the sum of the resistance (R) and reactance (X), where reactance is given by the formula:
X = 2 * π * f * L - 1 / (2 * π * f * C)
where f is the frequency.
Plugging in the given values:
R = 15 Ω
L = 4H
C = 25 × 10^(-6) F
f = 15.93 Hz
X = 2 * π * 15.93 * 4 - 1 / (2 * π * 15.93 * 25 × 10^(-6))
= 100.48 - 79572.28
≈ -79471.80 Ω
The total impedance is given by:
Z = square root (R^2 + X^2)
= square root (15^2 + (-79471.80)^2)
= square root (225 + 6314220327.24)
≈ square root (6314220552.24)
≈ 79471.81 Ω
Now, using Ohm's Law:
current (I) = voltage (V) / impedance (Z)
= 230 V / 79471.81 Ω
≈ 0.0029 A
Therefore, the current is approximately 0.0029 A.
c. To calculate the voltage drop across the inductance, use the formula:
voltage drop across inductance (V_L) = I * X
where I is the current and X is the reactance.
Plugging in the given values:
I = 0.0029 A
X = -79471.80 Ω
V_L = 0.0029 * -79471.80
≈ -229.94 V
Therefore, the voltage drop across the inductance is approximately -229.94 V.
To solve the given problem, we need to use basic electrical formulas and relationships. Here's how you can calculate the frequency, current, and voltage drop across the inductance:
a. Frequency (f):
To calculate the frequency, we'll use the formula:
f = 1 / (2π√(LC))
Where:
L = Inductance (4H)
C = Capacitance (25μF)
First, let's convert the capacitance from microfarads (μF) to farads (F):
25μF = 25 × 10^(-6) F
Now, we can substitute the values into the formula and calculate the frequency:
f = 1 / (2π√(4 × 25 × 10^(-6)))
Simplify the equation and perform the calculations to find the value of f.
b. Current (I):
To calculate the current, we'll use Ohm's Law:
I = V / Z
Where:
V = Voltage (230V)
Z = Impedance
To calculate the impedance, we need to find the total impedance of the resistor (15Ω) and the reactance of the inductor (XL) and the capacitor (XC). The reactance can be calculated using the formulas:
XL = 2πfL
XC = 1 / (2πfC)
Once we have XL and XC, we can find the Z by using the formula for impedance in a series circuit:
Z = √(R^2 + (XL - XC)^2)
Now, substitute the given values into these formulas to find the reactances and impedance and then use Ohm's Law to calculate the current.
c. Voltage drop across inductance (VL):
The voltage drop across the inductance can be calculated using the formula:
VL = XL × I
Substitute the calculated value of XL and I into the formula to find the voltage drop across the inductance.
Remember, using a scientific calculator or an online calculator will help you with the calculations involving the trigonometric functions and complex numbers.