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Comparative Study Of Various Alternating Current Circuits

Comparative Study Of Various Alternating Current Circuits :- In this article, we will carry out a comparative study of various AC circuits in different combinations.

(a) Comparative study of R-L, R-C, and L-C circuits in series.

(b) Comparative study of series R-L-C and parallel R-L-C circuits

In row (5) of the above table,

$\displaystyle {{I}_{0}}=\sqrt{I_{0R}^{2}+{{\left( {{I}_{0L}}-{{I}_{0C}} \right)}^{2}}}$

$\displaystyle \Rightarrow I_{0}^{2}=I_{0R}^{2}+{{\left( {{I}_{0L}}-{{I}_{0C}} \right)}^{2}}$

$\displaystyle \Rightarrow {{\left( \frac{{{E}_{0}}}{Z} \right)}^{2}}={{\left( \frac{{{E}_{0}}}{R} \right)}^{2}}+{{\left( \frac{{{E}_{0}}}{{{X}_{L}}}-\frac{{{E}_{0}}}{{{X}_{C}}} \right)}^{2}}$

$\displaystyle \Rightarrow {{\left( \frac{1}{Z} \right)}^{2}}={{\left( \frac{1}{R} \right)}^{2}}+{{\left( \frac{1}{{{X}_{L}}}-\frac{1}{{{X}_{C}}} \right)}^{2}}$

$\displaystyle \Rightarrow {{\left( Y \right)}^{2}}={{\left( G \right)}^{2}}+{{\left( {{S}_{L}}-{{S}_{C}} \right)}^{2}}$

$\displaystyle \therefore Y=\sqrt{{{\left( G \right)}^{2}}+{{\left( {{S}_{L}}-{{S}_{C}} \right)}^{2}}}$

Example 1.

Find the impedance of the given circuit.

Solution :

$\displaystyle Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$

$\displaystyle \Rightarrow Z=\sqrt{{{4}^{2}}+{{(9-6)}^{2}}}$

$\displaystyle \therefore Z=\sqrt{{{4}^{2}}+{{3}^{2}}}=\sqrt{25}=5\Omega$

Example 2.

Find the impedance of the given circuit.

Solution :

$\displaystyle Y=\sqrt{{{G}^{2}}+{{\left( {{S}_{L}}-{{S}_{C}} \right)}^{2}}}=\sqrt{{{\left( \frac{1}{3} \right)}^{2}}+{{\left( \frac{1}{6}-\frac{1}{3} \right)}^{2}}}$

$\displaystyle Y=\frac{\sqrt{2}}{6}{{\Omega }^{-1}}$

$\displaystyle \therefore Z=\frac{1}{Y}=\frac{6}{\sqrt{2}}\Omega$

Example 3.

In the given circuit, determine the reading of the AC ammeter, the potential difference across the resistor & the capacitor.

Solution :

$\displaystyle Z=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}}=\sqrt{{{\left( 10 \right)}^{2}}+{{\left( 20-10 \right)}^{2}}}=10\sqrt{2}\Omega$

$\displaystyle {{I}_{0}}=\frac{{{V}_{0}}}{Z}=\frac{100}{10\sqrt{2}}=\frac{10}{\sqrt{2}}Amp.$

Since the ammeter measures the RMS value of the current, therefore the reading of an AC ammeter is,

$\displaystyle {{I}_{rms}}=\frac{{{I}_{0}}}{\sqrt{\text{2}}}=\frac{\left( {}^{10}\!\!\diagup\!\!{}_{\sqrt{\text{2}}}\; \right)}{\sqrt{\text{2}}}=5Amp.$

Potential difference across the resistor and capacitor,

$\displaystyle {{V}_{0R}}=5\times 10=50\text{ }Volt$

$\displaystyle {{V}_{0C}}=5\times 10=50\text{ }Volt$

Example 4.

In an RLC circuit, R = 300 Ω, C = 20μF , L = 1.0 H, alternating source E_rms = 50 V and frequency (ν) = (50/π) Hz are connected in series. Find the value of RMS current in the circuit ?

Solution :

$\displaystyle {{I}_{rms}}=\frac{{{E}_{rms}}}{Z}=\frac{{{E}_{rms}}}{\sqrt{{{R}^{2}}+{{\left[ \omega L-\frac{1}{\omega C} \right]}^{2}}}}=\frac{50}{\sqrt{{{300}^{2}}+{{\left[ 2\pi \times \frac{50}{\pi }\times 1-\frac{1}{20\times {{10}^{-6}}\times 2\pi \times \frac{50}{\pi }} \right]}^{2}}}}$

$\displaystyle \Rightarrow {{I}_{rms}}=\frac{50}{\sqrt{{{(300)}^{2}}+{{\left[ 100-\frac{{{10}^{3}}}{2} \right]}^{2}}}}=\frac{50}{100\sqrt{9+16}}=\frac{1}{10}=0.1A$

Example 5.

If X_L = 50 Ω and X_C = 40 Ω, find the effective (RMS) value of current in the given circuit.

Solution :

$\displaystyle Z={{X}_{L}}-{{X}_{C}}=50-40=10\Omega$

$\displaystyle {{I}_{0}}=\frac{{{V}_{0}}}{Z}=\frac{40}{10}=4Amp.$

Hence, the effective value of the current (RMS value),

$\displaystyle {{I}_{rms}}=\frac{4}{\sqrt{2}}=2\sqrt{2}Amp.$

Note :-

Here X_L > X_C, therefore the current will lag the voltage by 90°.

Example 6.

Determine the potential difference across the inductor in the given circuit.

Solution :

$\displaystyle \because E_{0}^{2}=V_{0R}^{2}+V_{0L}^{2}$

$\displaystyle \Rightarrow E_{rms}^{2}=\left( {{V}_{rms}} \right)_{R}^{2}+\left( {{V}_{rms}} \right)_{L}^{2}$

$\displaystyle \Rightarrow \left( {{V}_{rms}} \right)_{L}^{2}=E_{rms}^{2}-\left( {{V}_{rms}} \right)_{R}^{2}$

$\displaystyle \therefore {{\left( {{V}_{rms}} \right)}_{L}}=\sqrt{E_{rms}^{2}-\left( {{V}_{rms}} \right)_{R}^{2}}=\sqrt{{{100}^{2}}-{{60}^{2}}}=\sqrt{6400}=80Volts$

Example 7.

Find the impedance of the given circuits :

Solution :

(i) This is a parallel circuit, so first we calculate the value of the admittance (Y) :

$\displaystyle Y={{S}_{L}}-{{S}_{C}}=\frac{1}{120}-\frac{1}{240}=+\frac{1}{240}$

$\displaystyle \Rightarrow Z=240\Omega$

(ii)

$\displaystyle E_{rms}^{2}=\left( {{V}_{rms}} \right)_{R}^{2}+\left( {{V}_{rms}} \right)_{L}^{2}$

$\displaystyle \Rightarrow {{E}_{rms}}=\sqrt{\left( {{V}_{rms}} \right)_{R}^{2}+\left( {{V}_{rms}} \right)_{L}^{2}}=\sqrt{{{5}^{2}}+{{12}^{2}}}=\sqrt{169}=13Volts$

$\displaystyle \therefore Z=\frac{{{E}_{rms}}}{{{I}_{rms}}}=\frac{13}{2}=6.5\Omega$

(iii) Given :

$\displaystyle R=3\Omega$

$\displaystyle {{X}_{L}}=\omega L=1\times 1=1\Omega$

$\displaystyle {{X}_{C}}=\frac{1}{\omega C}=\frac{1}{\left( 0.2 \right)\times \left( 1 \right)}=5\Omega$

Therefore

$\displaystyle Z=\sqrt{{{R}^{2}}+{{\left( {{X}_{L}}-{{X}_{C}} \right)}^{2}}}=\sqrt{{{3}^{2}}+{{\left( 1-5 \right)}^{2}}}=\sqrt{9+16}$

$\displaystyle \therefore Z=5\Omega$

Example 8.

When an alternating voltage of 220 V is applied across the terminals of device X, a current of 0.5 A flows in the circuit and it is in phase with the applied voltage. When the same voltage is applied across the terminals of another device Y, the same current again flows in the circuit, but it leads the applied voltage by π/2 radians.

(a) Name the devices X and Y.
(b) Now if the same voltage is applied to the series combination of X and Y, then calculate the current flowing in the circuit.

Solution :

(a) X is a resistor and Y is a capacitor.

(b) Initially, the current in both the devices is, therefore

$\displaystyle R={{X}_{C}}=\frac{220}{0.5}=440\Omega$