Qstnid -Iv-14-in A Series Lcr Circuit With An Ac Source, R = 300 Ω, C = 20 Μf, L = 1.0 Henry, Ε<sub>rms</sub> = 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. (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. (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. - 39: Alternating Current

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NEET-XII-Physics

39: Alternating Current

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Qstn# iv-14 Prvs-Qstn Next-Qstn

#14
In a series LCR circuit with an AC source, R = 300 Ω, C = 20 μF, L = 1.0 henry, ε_rms = 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. (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. (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.
Ans : Given:
Resistance in series LCR circuit, R = 300 Ω
Capacitance in series LCR circuit, C = 20 μF= 20 × 10^-6 F
Inductance in series LCR circuit, L = 1 Henry
RMS value of voltage, ε_rms = 50 V
Frequency of source, f = 50/`` \,\mathrm{\,\pi \,}`` Hz
Reactance of the inductor (X_L) is given by,
X_L= `` \omega L`` = 2`` \,\mathrm{\,\pi \,}``fL
`` \Rightarrow ``X_L = 2`` \times \,\mathrm{\,\pi \,}\times \frac{50}{\,\mathrm{\,\pi \,}}\times 1`` = 100 `` \,\mathrm{\,\Omega \,}``
Reactance of the capacitance `` \left({X}_{C}\right)`` is given by,
`` {X}_{C}=\frac{1}{\omega C}`` = `` \frac{1}{2\,\mathrm{\,\pi \,}fC}``
`` \Rightarrow ``X_C = `` \frac{1}{2\,\mathrm{\,\pi \,}\times {\displaystyle \frac{50}{\,\mathrm{\,\pi \,}}}\times 20\times {10}^{-6}}``
`` \Rightarrow `` X_C = `` 500`` `` \,\mathrm{\,\Omega \,}`` (a) Impedance of an LCR circuit `` \left(Z\right)`` is given by,
`` Z=\sqrt{{R}^{2}+{\left({X}_{\,\mathrm{\,C\,}}-{X}_{\,\mathrm{\,L\,}}\right)}^{2}}``
`` \Rightarrow Z=\sqrt{{\left(300\right)}^{2}+{\left(500-100\right)}^{2}}``
`` \Rightarrow Z=\sqrt{{\left(300\right)}^{2}+{\left(400\right)}^{2}}``
`` \Rightarrow Z=500``
`` ``
RMS value of current `` \left({I}_{rms}\right)`` is given by,
`` {I}_{rms}=\frac{{\epsilon }_{rms}}{Z}``
`` \Rightarrow ``I_rms = `` \frac{50}{500}``
`` \Rightarrow ``I_rms = `` 0.1\,\mathrm{\,A\,}`` (b) Potential across the capacitor `` \left({V}_{C}\right)`` is given by,
V_C = I_rms × X_C
`` \Rightarrow ``V_C = 0.1 × 500 = 50 V
Potential difference across the resistor `` \left({V}_{R}\right)`` is given by,
V_{R =}I_rms × R
`` \Rightarrow ``V_R = 0.1 × 300 = 30 V
Potential difference across the inductor `` \left({V}_{L}\right)`` is given by,
V_L = I_rms × X_L
`` \Rightarrow `` V_L= 0.1 × 100 = 10 V
R.M.S potential = 50 V
Net sum of all the potential drops = 50 V + 30 V + 10 V = 90 V
Sum of the potential drops > RMS potential applied
Page No 330: (a) Impedance of an LCR circuit `` \left(Z\right)`` is given by,
`` Z=\sqrt{{R}^{2}+{\left({X}_{\,\mathrm{\,C\,}}-{X}_{\,\mathrm{\,L\,}}\right)}^{2}}``
`` \Rightarrow Z=\sqrt{{\left(300\right)}^{2}+{\left(500-100\right)}^{2}}``
`` \Rightarrow Z=\sqrt{{\left(300\right)}^{2}+{\left(400\right)}^{2}}``
`` \Rightarrow Z=500``
`` ``
RMS value of current `` \left({I}_{rms}\right)`` is given by,
`` {I}_{rms}=\frac{{\epsilon }_{rms}}{Z}``
`` \Rightarrow ``I_rms = `` \frac{50}{500}``
`` \Rightarrow ``I_rms = `` 0.1\,\mathrm{\,A\,}`` (b) Potential across the capacitor `` \left({V}_{C}\right)`` is given by,
V_C = I_rms × X_C
`` \Rightarrow ``V_C = 0.1 × 500 = 50 V
Potential difference across the resistor `` \left({V}_{R}\right)`` is given by,
V_{R =}I_rms × R
`` \Rightarrow ``V_R = 0.1 × 300 = 30 V
Potential difference across the inductor `` \left({V}_{L}\right)`` is given by,
V_L = I_rms × X_L
`` \Rightarrow `` V_L= 0.1 × 100 = 10 V
R.M.S potential = 50 V
Net sum of all the potential drops = 50 V + 30 V + 10 V = 90 V
Sum of the potential drops > RMS potential applied
Page No 330: (a) Impedance of an LCR circuit `` \left(Z\right)`` is given by,
`` Z=\sqrt{{R}^{2}+{\left({X}_{\,\mathrm{\,C\,}}-{X}_{\,\mathrm{\,L\,}}\right)}^{2}}``
`` \Rightarrow Z=\sqrt{{\left(300\right)}^{2}+{\left(500-100\right)}^{2}}``
`` \Rightarrow Z=\sqrt{{\left(300\right)}^{2}+{\left(400\right)}^{2}}``
`` \Rightarrow Z=500``
`` ``
RMS value of current `` \left({I}_{rms}\right)`` is given by,
`` {I}_{rms}=\frac{{\epsilon }_{rms}}{Z}``
`` \Rightarrow ``I_rms = `` \frac{50}{500}``
`` \Rightarrow ``I_rms = `` 0.1\,\mathrm{\,A\,}`` (b) Potential across the capacitor `` \left({V}_{C}\right)`` is given by,
V_C = I_rms × X_C
`` \Rightarrow ``V_C = 0.1 × 500 = 50 V
Potential difference across the resistor `` \left({V}_{R}\right)`` is given by,
V_{R =}I_rms × R
`` \Rightarrow ``V_R = 0.1 × 300 = 30 V
Potential difference across the inductor `` \left({V}_{L}\right)`` is given by,
V_L = I_rms × X_L
`` \Rightarrow `` V_L= 0.1 × 100 = 10 V
R.M.S potential = 50 V
Net sum of all the potential drops = 50 V + 30 V + 10 V = 90 V
Sum of the potential drops > RMS potential applied
Page No 330:

#14-a
the rms current in the circuit and
Ans : Impedance of an LCR circuit `` \left(Z\right)`` is given by,
`` Z=\sqrt{{R}^{2}+{\left({X}_{\,\mathrm{\,C\,}}-{X}_{\,\mathrm{\,L\,}}\right)}^{2}}``
`` \Rightarrow Z=\sqrt{{\left(300\right)}^{2}+{\left(500-100\right)}^{2}}``
`` \Rightarrow Z=\sqrt{{\left(300\right)}^{2}+{\left(400\right)}^{2}}``
`` \Rightarrow Z=500``
`` ``
RMS value of current `` \left({I}_{rms}\right)`` is given by,
`` {I}_{rms}=\frac{{\epsilon }_{rms}}{Z}``
`` \Rightarrow ``I_rms = `` \frac{50}{500}``
`` \Rightarrow ``I_rms = `` 0.1\,\mathrm{\,A\,}``

#14-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.
Ans : Potential across the capacitor `` \left({V}_{C}\right)`` is given by,
V_C = I_rms × X_C
`` \Rightarrow ``V_C = 0.1 × 500 = 50 V
Potential difference across the resistor `` \left({V}_{R}\right)`` is given by,
V_{R =}I_rms × R
`` \Rightarrow ``V_R = 0.1 × 300 = 30 V
Potential difference across the inductor `` \left({V}_{L}\right)`` is given by,
V_L = I_rms × X_L
`` \Rightarrow `` V_L= 0.1 × 100 = 10 V
R.M.S potential = 50 V
Net sum of all the potential drops = 50 V + 30 V + 10 V = 90 V
Sum of the potential drops > RMS potential applied
Page No 330: