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The three impedances are connected in parallel across a 240v 50hz supply i) a resistance of 300 ohms ii)a coil of inductance 130mh and 55 ohm resistance iii)a 170 ohm resistance in series with a 25 uf capacitor. calculate circuit current through each impedance and total power comsumption of system?
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The three impedances are connected in parallel across a 240v 50hz supp...
Solution:

Calculation of Circuit Current through Each Impedance:

To calculate the circuit current through each impedance, we need to find the impedance of each component using the formulas:

Impedance of a Resistor = R
Impedance of an Inductor = jωL
Impedance of a Capacitor = 1/jωC

Where R is the resistance, L is the inductance, C is the capacitance, ω is the angular frequency (2πf), and j is the imaginary unit.

i) Impedance of Resistance:

The impedance of resistance is equal to its resistance, which is 300 ohms.

ii) Impedance of Inductor:

The impedance of an inductor is given by:

jωL = j(2πf)L = j(2π×50)(130×10^-3) = j40.84 ohms

Where f is the frequency and L is the inductance.

iii) Impedance of Capacitor:

The impedance of a capacitor is given by:

1/jωC = -j/(2πfC) = -j/(2π×50×25×10^-6) = -j127.32 ohms

Where C is the capacitance.

Now, we can calculate the total impedance of the circuit by adding the impedances of all three components in parallel.

Total Impedance = (1/R + 1/jωL + 1/jωC)^-1

= (1/300 + 1/j40.84 + 1/-j127.32)^-1

= 51.06 - j31.86 ohms

The circuit current can be calculated using Ohm's Law:

I = V/Z

Where V is the voltage and Z is the impedance.

i) Current through Resistance:

I1 = V/R = 240/300 = 0.8 A

ii) Current through Inductor:

I2 = V/ZL = 240/(51.06 - j31.86) = 3.08∠-33.8° A

iii) Current through Capacitor:

I3 = V/ZC = 240/(-j127.32) = 1.89∠90° A

Calculation of Total Power Consumption:

The total power consumption of the system can be calculated using the formula:

P = VI cos(φ)

Where V is the voltage, I is the current, and φ is the phase angle.

The total power factor can be calculated using the formula:

cos(φ) = Re{S}/|S|

Where S is the complex power and Re{S} is the real part of S.

The complex power can be calculated using the formula:

S = VI*

Where V* is the complex conjugate of V.

The total power consumption is:

P = VI cos(φ) = Re{S} = 240×(0.8 + 3.08∠-33.8° + 1.89∠90°) = 523.2 W

The total complex power is:

S = VI* = 240×(0.8 - j3.08∠33.8° + j1.89∠-90°) = 523.2 - j311.5 VA

The total power factor is:

cos(φ) = Re{
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The three impedances are connected in parallel across a 240v 50hz supply i) a resistance of 300 ohms ii)a coil of inductance 130mh and 55 ohm resistance iii)a 170 ohm resistance in series with a 25 uf capacitor. calculate circuit current through each impedance and total power comsumption of system?
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The three impedances are connected in parallel across a 240v 50hz supply i) a resistance of 300 ohms ii)a coil of inductance 130mh and 55 ohm resistance iii)a 170 ohm resistance in series with a 25 uf capacitor. calculate circuit current through each impedance and total power comsumption of system? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The three impedances are connected in parallel across a 240v 50hz supply i) a resistance of 300 ohms ii)a coil of inductance 130mh and 55 ohm resistance iii)a 170 ohm resistance in series with a 25 uf capacitor. calculate circuit current through each impedance and total power comsumption of system? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The three impedances are connected in parallel across a 240v 50hz supply i) a resistance of 300 ohms ii)a coil of inductance 130mh and 55 ohm resistance iii)a 170 ohm resistance in series with a 25 uf capacitor. calculate circuit current through each impedance and total power comsumption of system?.
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