Question

An a.c source of 200V, 50Hz connected across a 400Ω resistor and an inductor

of 3/ H in series. Calculate reactance, impendence, current in the coil.​

Answer 1

Given:

An a.c source of 200V, 50Hz connected across a 400Ω resistor and an inductor of 3/ H in series.

To find:

• Reactance
• Impedance
• Current in coil

Calculation:

Inductor reactance is given as :

$\therefore \: X_{L} = \omega \times L$

$\implies \: X_{L} = (2\pi f) \times L$

$\implies \: X_{L} = (2\pi \times 50) \times 3$

$\implies \: X_{L} = 942 \: ohm$

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Impedance is calculated as follows:

$\therefore \: Z = \sqrt{{X_{L}}^{2} + {R}^{2} }$

$\implies \: Z = \sqrt{ {(942)}^{2} + {(400)}^{2} }$

$\implies \: Z = \sqrt{ 1047364 }$

$\implies \: Z = 1023.4 \: ohm$

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Now, current:

$\therefore \: i = \dfrac{ voltage_{rms}}{Z}$

$\implies\: i = \dfrac{200 \sqrt{2} }{1023.4}$

$\implies\: i = 0.27 \: amp$

Hope It Helps.

Answer 2

$\huge \fcolorbox{lime}{aqua}{Solution -:}$

The frequency of AC source = F =50Hz

$\sf{Thus \: the \: impedance \: of \: the \: inductor,}$

$\sf{XL=ωL=(2πf)L=20πΩ}$

$\sf{Resistance \: of \: the \: resistor =20Ω}$

$\sf{Thus, \: the \: net \: impedance \: of \: the \: circuit =}$

$\sf\sqrt{XL²+R²} =65.94Ω$

$\sf{Thus, \: the \: current \: in \: the \: circuit =\frac{V}{X}} =$

$\sf{\frac{220V}{65.94Ω}=3.33A}$