Question

A 230v, 50hz voltage is applied to a coil having L=5H and R=2 ohm in series with a capacitance c .what must be the value of c in order to have a potential difference of 250v across the coil

Answer 1

Answer:

The value of the capacitance of the capacitor is 25.79μF.

Explanation:

Given a series LCR circuit.

Frequency of the circuit, f = 50Hz

Voltage across the coil, V = 230V

The inductance of the inductor, L = 5H

The resistance of the resistor, R = 2Ω

Potential difference across the coil, V = 250V

The inductive reactance is given by

$X_L=\omega L=2\pi fL$      

Substituting the values of frequency and inductance, the inductive reactance is  

$X_L=2\pi fL=2\times 3.14\times 50\times 5= 1570\Omega$

Impedance across the coil is given by

$Z^{2} =R^{2} +X_{L} ^{2}$

Substituting the resistance and inductive reactance, the impedance across the coil is

$Z^{2} =2^{2} +1570^{2}=4+2464900=2464904$

$\implies Z=\sqrt{2464904} =1570\Omega$

Current through the coil is

$I=\dfrac{V}{Z} =\dfrac{250}{1570}=0.159A$

Impedance across the circuit,  

$Z=\dfrac{V}{I} =\dfrac{230}{0.159}=1446.54\Omega$

Impedance across the circuit is also given by

$Z^{2} =R^{2} +(X_L-X_C)^{2}$

$(1446.54)^{2} =2^{2} +(1570-X_C)^{2}$

$\implies (1570-X_C)^{2}=2092480.51 -4$

$\implies (1570-X_C)^{2}=2092478.51$

$\implies 1570-X_C=\sqrt{2092478.51} =1446.54$

$\implies X_C=1570-1446.54=123.46$

Capacitive reactance, $X_C=123.46\Omega$

The capacitance of the capacitor is given by

$C=\dfrac{1}{\omega X_C} =\dfrac{1}{2\pi f X_C}$

   $=\dfrac{1}{2\times 3.14\times 50\times 123.46}=\dfrac{1}{38766.44}$

   $\approx2.579\times 10^{-5} =25.79\times 10^{-6}$

   $= 25.79\mu F$

Therefore, the value of the capacitance of the capacitor is 25.79μF.

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Answer 2

Answer:

Explanation:

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