Question

A coil of inductance 80 mH and resistance 60 ohm is connected to a 200 V, 100 Hz supply
Calculate the circuit impedance and the current taken from the supply.​

Answer 1

$\Large\bf{\color{indigo}GiVeN,}$

• A coil of inductance [L] = 80 mH = 80 × 10⁻³ H

• Resistance [R] = 60 Ω

• Voltage [Vᵣₘₛ] = 200 V

• Frequency [f] = 100 Hz

$\Large\bf{\color{coral}To\:FiNd,}$

1. The circuit impedance.
2. The current taken from the supply.

$\Large\bf{\color{lime}CaLcUlAtIoN,}$

$\bf\blue{We\:know\:that,}$

$\red\bigstar\:\:\bf\orange{Inductive\:reactance\:(X_L)\:=\:\omega\:L\:}$

$:\implies\:\:\bf{X_L\:=\:2\:\pi\:f\:L\:}$

$:\implies\:\:\bf{X_L\:=\:2\times{3.14}\times{100}\times{80\times{10^{-3}}}\:}$

$:\implies\:\:\bf{X_L\:=\:50240\times{10^{-3}}\:}$

$:\implies\:\:\bf\green{X_L\:=\:50.24\:\Omega\:}$

$\bf\purple{We\:know\:that,}$

$\pink\bigstar\:\:\bf\blue{Impedance\:(Z)\:=\:\sqrt{R^2\:+\:X_L^2}\:}$

$:\implies\:\:\bf{Z\:=\:\sqrt{(60)^2\:+\:(50.24)^2}\:}$

$:\implies\:\:\bf{Z\:=\:\sqrt{3600\:+\:2524.05}\:}$

$:\implies\:\:\bf{Z\:=\:\sqrt{6124.05}\:}$

$:\implies\:\:\bf{\color{olive}Z\:=\:78.25\:\Omega}$

$\Large\bold\therefore$ The circuit impedance is 78.25 Ω.

$\bf\green{We\:know\:that,}$

$\purple\bigstar\:\:\bf{\color{peru}Current\:(I)\:=\:\dfrac{Voltage}{Resistance}\:}$

$:\implies\:\:\bf{I\:=\:\dfrac{200}{60}\:}$

$:\implies\:\:\bf\orange{I\:=\:3.34\:A\:}$

$\Large\bold\therefore$ The current supplied is 3.34 A.

Answer 2

Given:-

A coil of inductance [L] = 80 mH = 80 × 10⁻³ H

Resistance [R] = 60 Ω

Voltage [Vᵣₘₛ] = 200 V

Frequency [f] = 100 Hz

To Find:-

The circuit impedance.

The current taken from the supply.

Solution:-

We have ,

$\sf\red{Inductive\:reactance\:(X_L)\:=\:\omega\:L\:}$

$\sf\to{X_L\:=\:2\:\pi\:f\:L\:}$

$\sf\to{X_L\:=\:2\times{3.14}\times{100}\times{80\times{10^{-3}}}\:}$

$\sf\to{X_L\:=\:50240\times{10^{-3}}\:}$

$\sf\to{X_L\:=\:50.24\:\omega\:}$

We have,

$\sf\blue{Impedance\:(Z)\:=\:\sqrt{R^2\:+\:X_L^2}\:}$

$\sf\to{Z\:=\:\sqrt{(60)^2\:+\:(50.24)^2}\:}$

$\sf\to{Z\:=\:\sqrt{3600\:+\:2524.05}\:}$

$\sf\to{Z\:=\:\sqrt{6124.05}\:}$

$\sf \to Z\:=\:78.25\:\omega$

$\sf\therefore\:The\: circuit\: impedance \:is \:78.25 Ω.$

We have ,

$\sf\green{Current\:(I)\:=\:\dfrac{Voltage}{Resistance}\:}$

$\to{I\:=\:\dfrac{200}{60}\:}$

$\sf\implies{\boxed{I\:=\:3.34\:A\:}}$

$\sf\therefore\: The\: current \:supplied\: is \:3.34 A.$