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,
Answers
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