Chemistry, asked by digital2940, 1 year ago

In a series LCR circuit with an AC source, R = 300 Ω, C = 20 μF, L = 1.0 henry, εrms = 50 V and ν = 50/π Hz. Find (a) the rms current in the circuit and (b) the rms potential difference across the capacitor, the resistor and the inductor. Note that the sum of the rms potential differences across the three elements is greater than the rms voltage of the source.

Answers

Answered by bhuvna789456
0

(a) The rms current in the circuit is 0.1 A

(b) The rms potential difference across the capacitor, the resistor and the inductor is 90 V

Explanation:

Resistance in LCR circuit sequence, R = 300 Ω

Potential on LCR series circuit, C = 20 μF= 20 × 10−6 F

Inductance in LCR circuit series, L = 1 Henry

RMS Voltage Value, εrms   = 50 V

Source Frequency, f =50/π  Hz

The inducer reactance (XL) is given by,

X_{L}=\omega L=2 \pi f L

X_{L}=2 \times \pi \times \frac{50}{\pi} \times 1

     = 100 Ω

Capacitance Reactance (X_C) is given through,

X_{C}=\frac{1}{\omega C}=\frac{1}{2 \pi f C}

X_{C}=\frac{1}{\left(2 \pi \times \frac{50}{\pi} \times 20 \times 10^{-6}\right)}

X_{C}=500 \Omega

(a) The rms current in the circuit :

Impedance of an LCR (Z) circuit is provided by,

z=\sqrt{R^{2}+\left(X_{C}-X_{L}\right)^{2}}  

z=\sqrt{(300)^{2}+(500-100)^{2}}

z=\sqrt{(300)^{2}+(400)^{2}}

z=500

The current RMS value (Irms) shall be given by,

I_{r m s}=\frac{\varepsilon_{r m s}}{Z}

I_{r m s}=\frac{50}{500}

I_{r m s}=0.1 A

(b) The rms potential difference across the capacitor, the resistor and the inductor:

The condenser potential (V_C) is given by,

v_{c}=I_{r m s} \times \mathrm{x}_{c}

V_{C}=0.1 \times 500=50 \mathrm{V}

Potential resistor difference (V_R) is given by,

\begin{array}{l}{V_{R}=I_{r m s} \times R} \\{V_{R}=0.1 \times 300=30 \mathrm{V}}\end{array}

Potential inducer difference (V_L) is given by,

V_{R}=I_{r m s} \times R

V_{R}=0.1 \times 300=30 \mathrm{V}

R.M.S potential = 50 V  

Total sum of all potential falls = 50 V + 30 V + 10 V = 90 V

Total potential falls  > use of RMS potential

Answered by ReRepeater
0

Answer:

Answer is here by attached  .hope you understand

Explanation:

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