Question

A 100 Omega resistasnce is connected in series with a 4 H inductor. The voltage across the resistor is V_R=(2.0V)sin(10^3 rad//s)t: (a) Find the expessinocircuit current (b) Find the inductive reactance (c) derive an expression for the voltage across the inductor,

Answer 1

Instantaneous current equation at time t is I(t) = (1/50√1601 ) sin(10³)t

Inductive reactance is X(l) = 4×10³ Ω

Voltage across inductor will be

V(L) = (80/√1601)cos(10³)t

•Resistance (R) = 100Ω

•inductor of inductance L = 4H

•Voltage source is V = 2sin(10³)t

Generally , Voltage source is of the form V = V•sinwt

V• is known as peak voltage

V• = 2V

w is known as time period of AC

cycle

w = 10³

•Inductive reactance of inductor is w times inductance i.e.

X(l) = wL

X(l) = 10³×4

X(l) = 4×10³ Ω

•Now reactance of whole circuit will be :

Z = √(X(l)² + R²)

Z = √[(4×10³)² + (100)²]

Z = √(16×10^6 + 10⁴)

Z = 100√(1600+1)

Z = 100√1601 Ω

•Instantaneous current equation will be :

I(t) = (V•/Z)sinwt

I(t) = (2/100√1601)sin(10³)t

I(t) = (1/50√1601 ) sin(10³)t

•Now , Voltage Across inductor will be

V(L) = Ldi/dt

I(t) = (1/50√1601 ) sin(10³)t

di/dt = [(1/50√1601 ) cos(10³)t] × 10³

di/dt = (20/√1601)cos(10³)t

also , inductance of inductor is 4H

•so, Voltage across inductor will be as follows :

V(L) = (80/√1601)cos(10³)t