Question

A coil of 0.01 henry inductance and 1 ohm resistance is connected to 200 Volt, 50 Hz AC supply.
1) Find the impedance of the circuit.
2) Find the time lag between max. alternating voltage and current.

Answer 1

1) Given,

Inductance$\left(L\right)=0.01H$

Resistance $\left(R\right)=1\Omega$

Voltage $\left(V\right)=200V$

Frequency$\left(f\right)=50Hz$.

Inductive reactance,

$X_L=2\pi fL=2\times 3.14\times 50\times 0.01=3.14\Omega$

Impedance of the R-L circuit is given by,

$Z=\sqrt{R^2+X_L^2}$

=$\sqrt{R^2+(2\pi fL{)}^2}$

=$\sqrt{1^2+(2\times 3.14\times 50\times 0.01{)}^2}$

$Z=\sqrt{10.86}=3.3\Omega$


2) Given,
Inductance $\left(L\right)=0.01H$

Resistance $\left(R\right)=1\Omega$

Voltage $\left(V\right)=200V$

Frequency $\left(f\right)=50Hz$

Inductive reactance, $X_L=2\pi fL=2\times 3.14\times 50\times 0.01=3.14\Omega$

Step 1: Find impedance.

Impedance of the R-L circuit is given by,

$Z=\sqrt{R^2+X_L^2}$

=$\sqrt{R^2+(2\pi fL{)}^2}$

=$\sqrt{1^2+(2\times 3.14\times 50\times 0.01{)}^2}$

$Z=\sqrt{10.86}=3.3\Omega$

Step 2: Find phase of the circuit.

$\tan \varphi =\frac{\Omega L}{R}=\frac{2\pi fL}{R}=\frac{2\times 3.14\times 50\times 0.01}{1}=3.14$

=${\tan }^{-1}\left(3.14\right)=72$

Phase difference$\left(\varphi \right)=\frac{72\times \pi }{180}rad$
.
Step 3: Find time lag.

Time lag between alternating voltage and current,

$\Delta t=\frac{\varphi }{\omega }=\frac{72\pi }{180\times 2\times 50}=\frac{1}{250}s=0.004s$

Final Answer: $0.004s$