Question

An inductive coil having negligible resistance
and 0.1H inductance is connected across 200V.
50HZ supply
Find 1) inductive reactance (2) rm.s value of
current (3) Power (4) Equation for voltage and
current
​

Answer 1

Answer:

The formula for calculating the inductive reactance of a coil is: inductive reactance, or XL, is the product of 2 times p (pi), or 6.28, the frequency of the ac current, in hertz, and the inductance of the coil, in henries. XL =2p x f x L.

However, the voltage drop across the inductor, VL will have a value equal to: Ve(-Rt/L). Then the voltage across the inductor, VL will have an initial value equal to the battery voltage at time t = 0 or when the switch is first closed and then decays exponentially to zero as represented in the above curves.

Answer 2

Given: Inductance of coil = 0.1H,

         Voltage = 200V

         Frequency = 50Hz

To find: (1) Inductive reactance

             (2) rms value of  current

             (3) Power

             (4) Equation for voltage and  current

​Solution:

(1) Inductive reactance is the opposition presented to alternating current by inductance.

XL = 2πfL

Where f is the frequency, L is the inductance, XL is the inductive reactance.

     = 2π(50)(0.1)

     = 31.4 Ω

(2) rms value of current is the current divided by √2.

Current in the circuit = Voltage/Resistance

                                  = 200/31.4 = 63.69 A

rms value of current = Current/√2

                                 = 63.69 /√2 A

                                 = 45.04 A

The rms value of current in circuit is 45.04 A

(3) Power in the circuit, P = VI ( Watts)

P = 200×63.69

 = 12738 Watt.

(4) Cuurent equation = I₀Sin(ωt)

                                    = 63.69Sin(2π50t)

Voltage equation = V₀Cos(ωt)

                              = V₀Cos(2πft)

                              = 200Cos(2π50t)