Question

an 8 resistance, 16 inductive reactance and an unknown capacitor are connected in series across a 100 v, 50 hz supply. the current drawn by the circuit is 12.5 a. the value of capacitance of the capacitor will be

Answer 1

Inductive Reactance of a coil depends on the frequency of the applied voltage as .... a) The voltage of the supply if the frequency is 50Hz

What inductance will give the same reactance as a capacitor of 2 µF when both are at 50 HzOptions are ⇒ (A) 5 H

Answer 2

Given:

Resistance $(R)$ = 8Ω

Inductive reactance $(X_{L} )$ = 16Ω

Potential difference $(V)$ $= 100 V$

Frequency $(f)$ $=50Hz$

Current $(I)$ $=12.5A$

To find:

Capacitance $(C)$ of the capacitor

Solution:

Step 1

We know, Impedance of a $LCR$ circuit is given by

$Z=\sqrt{R^{2}+(X_{L}- X_{C} )^{2} }$

And, applying Ohm,s law in $LCR$ circuit, we get,

$V=IR\\V=IZ$  $;$ $or$

$I=\frac{V}{Z}$

$I=\frac{V}{\sqrt{R^{2}+ (X_{L}- X_{C} )^{2} } }$

Substituting the given values in the equation, we get

$12.5=\frac{100}{\sqrt{(8)^{2}+(16-X_{C} )^{2} } }$

$\sqrt{64+(16-(X_{C} ))^{2} } =8 \\64+(16-X_{C} )^{2}=64$

Hence,

The value of Capacitive reactance in the given circuit is $X_{C}$ = 16Ω.

Step 2

We know, $X_{C}=\frac{1}{wC}$

Where, ω = 2πf

Substituting the given values in the equation, we get

$16=\frac{1}{2(3.14)(50)C}$

Hence,

$C=\frac{1}{16(314)} \\C=\frac{1}{5024}$ $;$ $or$

$C=200$ × $10^{-6}F$ $(approx)$

Final answer:

Hence, the value of capacitance in the given $LCR$ circuit is 200μF.