KCL/ KVL in AC Circuits MCQ Quiz in मल्याळम - Objective Question with Answer for KCL/ KVL in AC Circuits - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

When current source i(t) connected across an inductor(L), then the voltage across the inductor is given as

$V_L(t)=L\frac{di(t)}{dt}$

When voltage source v(t) connected across a capacitor(C), then the current across the capacitor is given as

$I_C(t)=C\frac{dv(t)}{dt}$

Calculation:

Given i(t) = (8e-250t - 4e-1000t) A for t ≥ 0, L = 0.5 mH

Then the voltage across the inductor is given as

$V_L(t)=L\frac{di\left(t\right)}{dt}=L(-2000e^{-250t}+4000e^{-1000t})$

At t = 0 its value is given as

V_L = 0.5 × 10^-3 × (- 2000 + 4000)

VL = 1 V

We use the following source transformation as:

 ⇒ 

 ⇒ 

The circuit becomes as:

Combining parallel current source

Transforming voltage source, we redraw the circuit as:

Z₁ = -j1 + 1Ω

$I_1=\frac{-j20}{-j1+1}=10-j10A$

Applying Current division, we get:

$I=\frac{2\parallel Z_1}{\left(2\parallel Z_1\right)+2}\left(I_1+20\mathrm{\angle }{0}^{\circ }\right)$

$2\parallel Z_1=\frac{4}{5}-j\frac{2}{5}$

$=\frac{\frac{4}{5}-\frac{2}{5}j}{\frac{4}{5}-\frac{2}{5}j+2}\left(10-j10+20\right)$                 

$=\frac{4-j2}{14-j2}\left(30-j10\right)$

$=8-j6A$

Concept:

Current through an open circuit or through a circuit with impedance tending to infinity is zero.

Calculation:

We are given a circuit shown below:

The impedance seen from the voltage source is Z_eq (say) which can be calculated as:

$Z_{eq}=\left[j\left(5-\frac{1}{\omega C}\right)||j\omega L\right]+50+j\omega L$

With $\omega =5000andL=1mH$ 

$Z_{eq}=\left[j\left(5-\frac{1}{\omega C}\right)||j5\right]+50+j5$  

$\Rightarrow Z_{eq}=-\frac{-\left(25-\frac{5}{\omega C}\right)}{\left(10-\frac{1}{\omega C}\right)j}+50+j5$

For I = 0, Z_eq → ∞ --- (1)

Z_eq = ∞, So $10-\frac{1}{\omega C}=0$

$10=\frac{1}{\omega C}\Rightarrow C=\frac{1}{5000\times 10}$

$C=20\mu F$

Calculation:

Let the voltage at point A be V_a and the voltage at point B be V_b as shown:

By voltage division rule,

$V_a=\frac{V_s\times \frac{1}{sC}}{(R+\frac{1}{sC})}$

Similarly, $V_b=\frac{V_s\times R}{R+\frac{1}{sC}}$

Now, V_o = V_a - V_b

$\Rightarrow V_o=\frac{V_s}{sC\left(\frac{sCR+1}{SC}\right)}-\frac{V_{s}R\left(sC\right)}{sCR+1}$

$\Rightarrow V_o=V_s(\frac{1}{1+sCR}-\frac{sCR}{1+sCR})$

In delta connection, V_L = V_ph = 400 Ω

$I_{ph}=\frac{V_{ph}}{R}=\frac{400}{400}=1A$

 I_Line = √3 I_phase = √3 × 1 = 1.732 A

Calculation of V_th:

The load should be open circuited as shown below.

So, V_th is the phase voltage in a 3 phase Balanced system

$V_{th}=\frac{{\mathrm{V}}_{\mathrm{L}}}{\sqrt{3}}=\frac{400}{\sqrt{3}}$

Calculation of Z_th:

After removing all the sources, all the resistors are connected in parallel.

${\mathrm{Z}}_{\mathrm{t}\mathrm{h}}=300300300=\frac{300}{3}=100\mathrm{\Omega }$

Now, the Thevenin equivalent circuit can be represented as given below.

Applying Voltage Division Rule,

${\mathrm{V}}_{\mathrm{x}}=\frac{100}{100+100}\cdot \frac{400}{\sqrt{3}}=\frac{200}{\sqrt{3}}=115.47\mathrm{V}$

Hence the r.m.s value of voltage across 100 Ω resistor is 115.47 V

Alternate method:

The given circuit can be redrawn as given below.

By applying the KVL in the first loop,

375 I₁ - 300 I₂ = 400∠0°

By applying the KVL in the second loop,

-300 I₁ - 600 I₂ = 400∠-120°

By solving the above two equations, we get

I₁ = 1.5396 ∠-30°

V = 75 I₁ = 115.47∠-30°

The voltage across 100 Ω resistor is 115.47 V.

By using source transformation technique,

$Z=\sqrt{R^2+{\left(\frac{1}{\omega C}\right)}^2}$

$=\sqrt{{\left(2\right)}^2+{\left(\frac{1}{0.5}\right)}^2}=2\sqrt{5}\mathrm{\Omega }$

$Z=2\sqrt{2}=\mathrm{\angle }-{45}^{\circ }$

$i\left(t\right)=\frac{2\sqrt{2}\sin \left(5t-{45}^{\circ }\right)}{2\sqrt{2}\mathrm{\angle }-{45}^{\circ }}=\sin 5tA$

$\frac{4}{3}=\frac{X_L-X_C}{R}$, $\frac{X_C}{R}=\frac{3}{4}$

Z = $\frac{V}{\mathrm{I}}=\frac{100}{5}=20=\sqrt{R^2+(X_L-X_C)2}$

400 = $R^2+\frac{16R^2}{9}=\frac{25R^2}{9}$

R = 12Ω

$\frac{4}{3}=\frac{X_L}{R}-\frac{X_C}{R}$ ⇒ $\frac{4}{3}=\frac{X_L}{R}-\frac{3}{4}$

$\frac{X_L}{R}=\frac{16+9}{12}=\frac{25}{12}$

X_L = $\frac{25}{12}\times 12=25$

X_L = ωL = 25

L = $\frac{1}{4}H$  Ans.

Explanation:

Waveform: The curve obtains by plotting the instantaneous value of any electrical quantity such as voltage, current, or power.

Cycle: One complete set of positive and negative values or maximum or minimum value of the alternating quantity is called a cycle.

Note: One-half cycle of the wave is called Alternation.

Amplitude: the maximum value of the positive or negative alternative quantity is called Amplitude.

• For sinusoidal current, the unit of the amplitude is amperes.
• For sinusoidal voltage, the unit of the amplitude is volts.
• It is also known as peak value or crest value.
• Peak-Peak Value = 2 × Amplitude
• In the given figure amplitude of the waveform is V_m.

Time period (T): It is the time required to complete one cycle. It measured in seconds.

Frequency: The number of cycles completed per second is called frequency.

$f=\frac{1}{T}$

It measured in Hz or radian/sec.

Concept:

In a series RLC circuit, under resonance condition, the inductive reactance is equal to the capacitive reactance and hence the circuit acts as pure resistive circuit.

Z = R + j(X_L - X_C)

At resonance, X_L = X_C

Z = R

The current flows in the circuit is:

$i\left(t\right)=\frac{V\left(t\right)}{Z}=\frac{V\left(t\right)}{R}$

Calculation:

V(t) = 100 √2 cos 1000t

R = 100 Ω, L = 100 mH, C = 10 μF, ω = 1000

X_L = ωL = 100 Ω

$X_C=\frac{1}{\omega C}=100\mathrm{\Omega }$

Z = 100 + j (100 - 100) = 100

$i\left(t\right)=\frac{V\left(t\right)}{Z}=\frac{\sqrt{2}100\mathrm{c}\mathrm{o}\mathrm{s}1000\mathrm{t}}{100}=\sqrt{2}\mathrm{c}\mathrm{o}\mathrm{s}1000\mathrm{t}$

$I_{rms}=\frac{\sqrt{2}}{\sqrt{2}}=1$