Question

A coil of 0.01H inductance and 1 ohm resistance is connected to 200 volt, 50 Hz ac supply. Find the impedance of the circuit and time lag between maximum alternating voltage and current

Answer 1

Explanation:

A coil of 0.01H inductance and 1 ohm resistance is connected to 200 volt, 50 Hz ac supply. Find the impedance of the circuit and time lag between maximum alternating voltage and current answer

Answer 2

$\begin{gathered}\Large\bf{\color{indigo}GiVeN,} \\ \end{gathered}$

• A coil of inductance [L] = 80 mH = 80 × 10⁻³ H
• Resistance [R] = 60 Ω
• Voltage [Vᵣₘₛ] = 200 V
• Frequency [f] = 100 Hz

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

• The circuit impedance.
• The current taken from the supply.

$\begin{gathered}\Large\bf{\color{lime}CaLcUlAtIoN,} \\ \end{gathered}$

$\begin{gathered}\bf\blue{We\:know\:that,} \\ \end{gathered}$

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

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

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

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

$\begin{gathered}:\implies\:\:\bf\green{X_L\:=\:50.24\:\text{\O}mega\:} \\ \end{gathered}$

$\begin{gathered}\bf\purple{We\:know\:that,} \\ \end{gathered}$

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

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

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

$\begin{gathered}:\implies\:\:\bf{Z\:=\:\sqrt{6124.05}\:} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{\color{olive}Z\:=\:78.25\:\text{\O}mega} \\ \end{gathered}$

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

$\begin{gathered}\bf\green{We\:know\:that,} \\ \end{gathered}$

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

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

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

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