NEET-XII-Physics
39: Alternating Current
- #18Figure (39-E1) shows a typical circuit for a low-pass filter. An AC input Vi = 10 mV is applied at the left end and the output V0 is received at the right end. Find the output voltage for ν = 10 k Hz, 1.0 MHz and 10.0 MHz. Note that as the frequency is increased the output decreases and, hence, the name low-pass filter.
Figure
Ans : Here,
Input voltage to the filter, Vi = 10 × 10-3 V
Resistance of the circuit, R = 1 × 103 Ω
Capacitance of the circuit, C = 10 × 10-9 F
(a) When frequency, f = 10 kHz
A low pass filter consists of resistance and capacitance. Voltage across the capacitor is taken as the output.
Capacitive reactance `` \left({X}_{C}\right)`` is given by,
`` {X}_{\,\mathrm{\,C\,}}=\frac{\mathit{1}}{\mathit{\omega }\mathit{C}}=\frac{1}{2\,\mathrm{\,\pi \,}fC}``
`` \Rightarrow {X}_{C}=\frac{1}{2\,\mathrm{\,\pi \,}\times 10\times {10}^{3}\times 10\times {10}^{-9}}``
`` \Rightarrow {X}_{C}=\frac{1}{2\,\mathrm{\,\pi \,}\times {10}^{-4}}``
`` \Rightarrow {X}_{C}=\frac{{10}^{4}}{2\,\mathrm{\,\pi \,}}=\frac{5000}{\,\mathrm{\,\pi \,}}\,\mathrm{\,\Omega \,}``
Net impedence of the resistance-capacitance circuit (Z) is given by,
`` Z=\sqrt{{R}^{2}+{X}_{\,\mathrm{\,C\,}}^{2}}``
`` \Rightarrow Z=\sqrt{\left(1+{10}^{3}\right)+{\left(5000/\,\mathrm{\,\pi \,}\right)}^{2}}``
`` \Rightarrow Z=\sqrt{{10}^{6}+{\left(5000/\,\mathrm{\,\pi \,}\right)}^{2}}``
`` ``
Current (I0) is given by,
`` {I}_{0}=\frac{{V}_{i}}{Z}``
`` \Rightarrow {I}_{0}=\frac{10\times {10}^{-3}}{\sqrt{{10}^{6}+{\left(5000/\,\mathrm{\,\pi \,}\right)}^{2}}}``
`` ``
Output across the capacitor `` \left({V}_{0}\right)`` is given by,
`` {V}_{0}=\frac{{10}^{2}}{\sqrt{{10}^{5}+{\left(50/\,\mathrm{\,\pi \,}\right)}^{2}}}\times \frac{500}{\,\mathrm{\,\pi \,}}``
`` \Rightarrow {V}_{0}=1.6124\,\mathrm{\,V\,}=1.6\,\mathrm{\,mV\,}``
(b)When frequency, f = 1 MHz = `` 1\times ``106 Hz
Capacitive reactance `` \left({X}_{C}\right)`` is given by,
`` {X}_{c}=\frac{1}{\omega C}``
`` \Rightarrow {X}_{C}=\frac{\mathit{1}}{\mathit{2}\pi fC}``
`` \Rightarrow {X}_{C}=\frac{1}{2\,\mathrm{\,\pi \,}\times {10}^{6}\times {10}^{-9}\times 10}``
`` \Rightarrow {X}_{C}=\frac{1}{2\,\mathrm{\,\pi \,}\times {10}^{-2}}``
`` \Rightarrow {X}_{C}=\frac{100}{2\,\mathrm{\,\pi \,}}``
`` \Rightarrow {X}_{C}=\frac{500}{\,\mathrm{\,\pi \,}}\,\mathrm{\,\Omega \,}``
`` \,\mathrm{\,Total\,}\,\mathrm{\,impedence\,}\left(Z\right)=\sqrt{{R}^{2}+{{\,\mathrm{\,X\,}}_{\,\mathrm{\,C\,}}}^{2}}``
`` \Rightarrow Z=\sqrt{{\left({10}^{3}\right)}^{2}+{\left(50/\,\mathrm{\,\pi \,}\right)}^{2}}``
`` \,\mathrm{\,Current\,}\left({I}_{0}\right)=\frac{{V}_{\mathit{1}}}{Z}``
`` \Rightarrow {I}_{0}=\frac{10\times {10}^{-3}}{\sqrt{{10}^{6}+{\left(50/\,\mathrm{\,\pi \,}\right)}^{2}}}``
`` \,\mathrm{\,Output\,}\,\mathrm{\,voltage\,}\left({V}_{0}\right)\mathit{=}{I}_{\mathit{0}}{X}_{C}``
`` \Rightarrow {V}_{0}=\frac{{10}^{-2}}{\sqrt{{10}^{5}+{\left(50/\,\mathrm{\,\pi \,}\right)}^{2}}}\times \frac{50}{\,\mathrm{\,\pi \,}}``
`` \Rightarrow {V}_{0}=0.16\,\mathrm{\,mV\,}``
(c) When frequency, f = 10 MHz = 10 × 106 Hz = 107 Hz
Capacitive reactance `` \left({X}_{C}\right)`` is given by,
`` {X}_{c}=\frac{1}{\omega C}``
`` \Rightarrow {X}_{C}=\frac{1}{2\pi fC}``
`` \Rightarrow {X}_{C}=\frac{1}{2\,\mathrm{\,\pi \,}\times {10}^{7}\times 10\times {10}^{-9}}``
`` \Rightarrow {X}_{C}=\frac{5}{\,\mathrm{\,\pi \,}}\,\mathrm{\,\Omega \,}``
`` \,\mathrm{\,Impedence\,}\left(Z\right)=\sqrt{{R}^{2}+{X}_{c}^{2}}``
`` \Rightarrow Z=\sqrt{{\left({10}^{3}\right)}^{2}+{\left(5/\,\mathrm{\,\pi \,}\right)}^{2}}``
`` \,\mathrm{\,Current\,}\left({I}_{0}\right)=\frac{{V}_{1}}{Z}``
`` \Rightarrow {I}_{0}=\frac{10\times {10}^{-3}}{\sqrt{{10}^{6}+{\left(5/\,\mathrm{\,\pi \,}\right)}^{2}}}``
`` {V}_{0}={I}_{0}{X}_{\,\mathrm{\,C\,}}``
`` \Rightarrow {V}_{0}=\frac{{10}^{-2}}{\sqrt{{10}^{6}+{\left(5/\,\mathrm{\,\pi \,}\right)}^{2}}}\times \frac{5}{\,\mathrm{\,\pi \,}}``
`` \Rightarrow {V}_{o}=16\,\mathrm{\,\mu V\,}``
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