SECTION -C A capacitor (C) and resistor (R) ...

CBSE Class 12 Physics 2019 Outside Delhi Set 1 Paper

Question : 13 of 27

Marks: +1, -0

SECTION -C

A capacitor

(C)

and resistor

(R)

are connected in series with an ac source of voltage of frequency

50\,\text{Hz}

. The potential difference across

C

and

R

are respectively

120\,\text{V}, 90\,\text{V}

, and the current in the circuit is

3\,\text{A}

. Calculate (i) the impedance of the circuit (ii) the value of the inductance, which when connected in series with

C

and

R

will make the power factor of the circuit unity.

The figure shows a series LCR circuit connected to a variable frequency

230\,\text{V}

source.

(a) Determine the source frequency which drives the circuit in resonance.

(b) Calculate the impedance of the circuit and amplitude of current at resonance.

Solution:

Ans. Calculation of impedance Calculation of inductance

(i)

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

\; R=\frac{V_R}{I_R}=30\,\Omega

\; X_C=\frac{V_C}{I_C}=\frac{120}{30}=40\,\Omega

\; Z=\sqrt{(30)^2+(40)^2}=50\,\Omega

\; X_C=X_L

(ii) As power factor

=1

\; 100 \pi L=40

\; L=\frac{2}{5\pi} \text{ henry}

Determining the source frequency

Calculating impedance

For showing potential drop across LC

(a)

\; \omega=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{5 \times 80 \times 10^{-6}}}

=\frac{1}{\sqrt{400 \times 10^{-6}}}

\; \omega=\frac{1000}{20}=50\,\text{Hz}

\; \text{(b)} \; Z=R=40\,\Omega

\omega=\frac{1000}{20}=50\,\text{Hz}

I_m^{\max}=\frac{230\sqrt{2}}{R}=\frac{230\sqrt{2}}{40}

=8.1\,\text{A}

V_c=I_m^{\max} X_c=\frac{230\sqrt{2}}{40} \times \frac{1}{\omega C}

=2033\,\text{volt}

V_L=I_m^{\max} X_L=\frac{230\sqrt{2}}{40} \times 2 \pi v L

=2033\,\text{volt}

(c)

V_C-V_L=0

Go to Question:

Prev Question

Next Question