Basic Electrical Engineering (3110005)

Question-2c

BE | Semester-1 Winter-2019 | 11-01-2020

Q2) (c)

Given Data

Supply Voltage,

V_{s} = 150 V

Resistance,

R = 8 Ω

Inductance,

L = 80 mH

Capacitance,

C = 100 μF

Task

Current,

I = ?

Power Factor,

p . f . = ?

Voltage drop across inductor,

{IX}_{L} = ?

Voltage drop across capacitor,

{IX}_{C} = ?

Calculations

The inductive reactance,

X_{L} = 2 π f L

\Rightarrow X_{L} = 2 \times 3.14 \times 50 \times 80 \times 10^{- 3}

\Rightarrow X_{L} = 25 . 12 Ω

The capacitive reactance,

X_{c} = \frac{1}{2 π f C}

\Rightarrow X_{c} = \frac{1}{2 \times 3.14 \times 50 \times 100 \times 10^{- 6}}

\Rightarrow X_{c} = 31 . 84 Ω

Since,

X_{C} = X_{L}

The Impedance,

Z = \sqrt{R^{2} + {(X_{c} - X_{L})}^{2}}

\Rightarrow Z = \sqrt{8^{2} + {(31.84 - 25.12)}^{2}}

\Rightarrow Z = \sqrt{8^{2} + {(6.72)}^{2}}

\Rightarrow Z = \sqrt{64 + 45.15}

\Rightarrow X_{c} = 10 . 44 Ω

RMS current,

\frac{V}{Z} = \frac{150}{10.44} = 14.36 A

Power Factor,

\frac{R}{Z} = \frac{8}{10.44} = 0.766

Voltage across the Inductor,

I X_{L} = 14.36 \times 25.12 = 360.72 V

Voltage across the Capacitor ,

I X_{C} = 14.36 \times 31.84 = 457.22 V