Question

An inductor of inductance l and resistor of resistance r are joined in series and connected by source of frequency power dissipated in the circuit is

Answer 1

Answer:

Given is series R – L circuit

Hence impedance = Z = √[R^2 + XL^2]

current I = [V / {|Z|}] = [V / √{R^2 + XL^2}]

power dissipated in the circuit is P = I2^R = [{V}/√{R^2 + XL^2}]^2 ∙ R

∴ P = [{V^2R}/{R^2 + XL^2}] = [{V^2R}/{R^2 + ω^2L^2}].

Answer 2

An inductor of inductance l and resistor of resistance r are joined in series and connected by source of frequency power dissipated in the circuit is

P = V^2R / (R^2 + w^2L^2)

• Given:
• Inductor and Resistor are in series
• Implies, the impedance
• Z = √(R^2 + XL^2)
• Current, I
• I = V / |Z|
• ⇒ I = V / √(R^2 + XL^2)
• Now, the power dissipated in the circuit, is given by,
• P = I^2R
• = [V / √(R^2 + XL^2)]^2 * R
• ⇒ P = V^2R / (R^2 + XL^2)
• ∴ P = V^2R / (R^2 + w^2L^2)