Question

A voltage v=100sin314t is applied to a circuit consisting of 25ohm resistor and an 80microfarad capacitor in series. Determine () Peak value of current (ii) Power factor (ii) Total power consumed by the circuit.​

Answer 1

Explanation:

Impedance Z=R2+XL2

                                =R2+(ωL)2

                                =102+(300×800×10−3)2

                        Z=240.2Ω

(ii) I0=ZV0=240.2200=0.832A

Answer 2

Answer:

Peak value of the current is 2.13A, Power factor is 53.1, total power consumed is 79.96W.

Explanation:

Given an RC circuit connected in series.

The ac voltage supplied to the circuit, $v=100sin314t$

The resistance of a resistor, $R=25\Omega$

The capacitance of a capacitor, $C=80\mu F=80\times 10^{-6} F$

(i) The general representation of an ac voltage is $v=v_0sin\omega t$

where $v_0$ is the peak voltage and $\omega$ is the angular frequency.

Comparing this with the given ac voltage, the peak voltage, $v_0=100V$

Comparing the given ac voltage with the standard form, the angular frequency, $\omega=314rad/s$

The capacitive reactance is given by

$X_c=\dfrac{1}{\omega C}$

    $=\dfrac{1}{314\times 80\times 10^{-6} }=3.98\times10^{-5}\times 10^{6} =39.8\Omega$

Impedance, Z is given by

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

   $=\sqrt{(25)^{2} +(39.8)^{2} } =\sqrt{625+1584.04 }=\sqrt{2209.04 }\approx 47\Omega$

The peak value of the current in the circuit is given by

$I_0=\dfrac{v_0}{Z}$

   $=\dfrac{100}{47} \approx 2.13A$

Therefore, the peak value of the current is 2.13A

(ii) Power factor is given by

$pf=cos\theta=\dfrac{R}{Z}$

               $=\dfrac{25}{47} =0.531$

or we can say the current is leading by 53.1%.

(iii) Total power consumed by the circuit is given by

$P=v_{rms}\times I_{rms} \times cos\theta$

   $=\dfrac{100 }{\sqrt{2} }\times \dfrac{2.13 }{\sqrt{2} }\times 0.531$

   $=0.707\big(100\times2.13\big)\times 0.531=0.707\times213\times 0.531\approx 79.96W$

Therefore, total power consumed is 79.96W.

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