In a LR circuit, R=10Ωand L=2H. If an alternating voltage of 120V and 60Hz is connected in this circuit, then the value current flowing in it will be ____________A

 By using Ohm’s law V=IR we can find the value of current flowing in the circuit

Here we have the value of potential difference 120V. In LR circuits, resistance is in the form of impedance. Therefore, impedance, Z= √R2+X2L

Z = √R2+(2πfL)2

(ii) Applying the given values in the above formula,

Z = √(10)+ (2×3.14×60×2)2

Z = √100+567912.96

Z = √568012.96−−−−−−−−√

∴ Z=753.66

(iii) In Ohm’s law V=IR

The current flowing in the circuit is

⇒ I = V/R

For LR circuit, the current flowing in the circuit is I = V/Z

→I = 120/753.66

∴I = 0.159A

Hence, I = 0.16A(approximately)

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