Question

A 200 V, 50 Hz, inductive circuit takes a current of 10A, lagging 30 degree. Find (i) the resistance (ii) reactance (iii) inductance of the coil.​

Answer 1

Answer:

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Explanation:

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Answer 2

Given:

• Voltage (V) = 200V
• Current (I) = 10A
• frequency (f) = 50Hz

To find:

• Resistance.
• Reactance.
• The inductance of the coil.

Solution:

• First we are finding the value of impedance(Z), Z = V/I = 200/10 = 20 ohm.
• To find resistance we use the below formula,
• R = Zcosθ = 20*cos(30°) = $20*\frac{\sqrt{3} }{2} = 10*{\sqrt{3} } = 17.32 ohm$
• Now Inductance is given by the formula, L = $\frac{X_L}{2*pi*f}$
• Finding the Reactance of the coil using,
• Where $X_L =$ Zsinθ = 20*(1/2) = 10 ohm
• So, L = $\frac{10}{2*3.14*50}$ = 0.031H(0.031 Henry)

Resistance = 20 ohm

Reactance = 10 ohm

Inductance = 0.031 H