Steady State Ac Analysis – GATE EE Electric CircuitsPractice Questions & PYQs

Case (i) $\mathbf{t} \lt 1$ At steady state, capacitor behaves as open circuit and inductor behaves as short circuit. Redraw the circuit: Using current division,

Case (ii) $\mathbf{t} \gt 0$ : Switch is opened. $\because \quad \mathrm{V}_{\mathrm{C}}\left(0^{-}\right)=\mathrm{V}_{\mathrm{C}}\left(0^{+}\right)=12 \mathrm{~V}$ and $\quad \mathrm{i}_{\mathrm{L}}\left(0^{-}\right)=\mathrm{i}_{\mathrm{L}}\left(0^{+}\right)=6 \mathrm{~A}$ Redraw the circuit $\mathrm{t}=0^{+}$ : Apply KVL in loop,

We have, overall Q-factor of given circuit is, $Q=\frac{Q_LQ_C}{Q_L+Q_C}=\frac{60 \times 240}{60+240}=48$

$Q=\frac{\omega _o}{BW}=\frac{f_o}{BW}=\frac{150 \times 10^3}{600}=250$

Let the time required by the voltage across the capacitor to drop to 1 V is $t_1,$

$V_2:\frac{\pi}{18}=\frac{180^{\circ}}{18}=10^{\circ}$ $V_3:\frac{\pi}{36}=\frac{180^{\circ}}{36}=5^{\circ}$ $V_2$ leads $V_1$ and $V_3$ , So, $V_2$ is a source, $V_1$ and $V_3$ are absorbing. Hence, $P_2 \gt 0$ and $P_1,P_3 \lt 0$

Given that, $v(t)=5-10 \cos (\omega t+60^{\circ})$ $i(t)=5+C \cos (\omega t-0^{\circ})$ $P_{req}=0$ $0=5 \times 5 +\frac{1}{2}[(-10)(X)\cos(60^{\circ})]$ $-25=\frac{1}{2}[(-10)(X)\cos(60^{\circ})]$ $X=10$

As per GATE Official answer key MTA (Marks to All)

The average power consumed by the load = $P=V_1I_1 \cos \phi$ $\;\;=\frac{100}{\sqrt{2}}\frac{10}{\sqrt{2}} \cos 60^{\circ}=250W$

If we observe the parallel LC combination, we get that at $\omega$ =1000 rad/sec the parallel LC is at resonance, thus it is open circuited. The circuit given in question can be redrawn as So, $I=\frac{10 \sin 1000t}{10}=\sin 100t$ So, peak value is 1 Amp.

Currently no explanation available

Given, total power dissipated in the circuit = 1kW =1000 Watt $\therefore \;\; 2^2 \times 1 +10^2 \times R=1000$ $R=\frac{998}{100}=9.98\Omega$ Also, voltage drop across R, $V_R=IR$ $=10 \times 9.98=99.8 volt$ Voltage drop across load, $V=200 volt =\sqrt{V_R^{2+}V_{X_L}^2}$ $\therefore$ voltage drop across inductor, $V_{X_L}=\sqrt{V^{2-}V_R^2}$ $\;\;=\sqrt{(200)^{2-}(99.8)^2}$ $\;\;=173.32 volt$ Now, $V_{X_L}=IX_L$ $X_L=\frac{V_{X_L}}{I}$ $X_L=\frac{173.32}{10}=17.332\Omega$

$i=5 \cos (100 \pi t +100^{\circ})A$ $\;\;=5\angle 100^{\circ}A$ $Z=(4-j3)\Omega$ $\;\;=5\angle -36.87^{\circ}\Omega$ $V=iZ=25\angle 63.13^{\circ}V$ The average power is $P=\frac{1}{2}V_mI_m \cos \phi$ $\;\;=\frac{1}{2} \times 25 \times 5 \times \cos (36.87^{\circ})$ $\;\;=\frac{1}{2} \times 25 \times 5 \times \frac{4}{5}=50W$ Alternate Method: $P=|I_{rms}|^2R$ $P=\left ( \frac{5}{\sqrt{2}} \right )^2 4=50W$

Given, $R_L= 5\Omega$ $\therefore \;\; \cos \phi =\frac{R_L}{|Z|}$ $\Rightarrow \;\;|Z|=\frac{5}{0.45}=11.11$ Power consumed by load, $P_L=\left ( \frac{V_3}{|Z|} \right )^2R_L$ $\;\;=\left ( \frac{136}{11.11} \right )^2 \times 5$ $\;\;=749.1\approx 750W$

$V_s=1 \sin t\equiv V_m \sin \omega t$ $V_m=1 V$ and $\omega =1$ rad/sec Impedance of the branch containing inductor and capacitor $Z=j(X_L-X_C)$ $\;\;=j\left ( \omega L-\frac{1}{\omega C} \right )$ $\;\;=j\left ( 1 \times 1 -\frac{1}{1 \times 1} \right )=0$ So, this branch is short circuit and the whole current flow through it $i(t)=\frac{1.0 \sin t}{1}=1.0 \sin t$ rms value of the current $=\frac{1}{\sqrt{2}}A$

Using KCL, $-I_s+I_{RL}+I_C=0$ $\Rightarrow \; I_C=I_s-I_{RL}$ $\;\;\;=\sqrt{2}\angle \frac{\pi}{4}-\sqrt{2}\angle -\frac{\pi}{4}$ $\;\;\;=2\angle 90^{\circ}=+j2A$

$V(t)=100\sqrt{2} \cos (100 \pi t)$ Voltage represented in phasor form: $V_{ph}=V_{rms}\angle \phi$ $V_{ph}=\frac{100\sqrt{2}}{\sqrt{2}}\angle 0^{\circ}$ $i(t)=10\sqrt{2} \sin \left ( 100 \pi t +\frac{\pi}{4} \right )$ $i(t)=10\sqrt{2} \cos \left ( 100 \pi t +\frac{\pi}{4}-\frac{\pi}{2} \right )$ $I_{ph}=\frac{10\sqrt{2}}{\sqrt{2}}\angle\left ( \frac{\pi}{4}-\frac{\pi}{2} \right )$ $\;\;=10\angle \left ( -\frac{\pi}{4} \right )A$

Power supplied by the source $=V_sI_s \cos \phi$ Where, $\phi =$ angle between $V_s$ and $I_s \;=\; \pi/4$ Inductor and capacitor do not consume power. Therefore, power dissipated in R = Power supplied by the source $P_R=V_SI_S \cos \phi$ $\;\;=1 \times \sqrt{2} \times \cos \frac{\pi}{4}$ $\;\;=\sqrt{2} \times \frac{1}{\sqrt{2}}=1 W$

Assuming resistance of the heater=R (i) When heater connected to 230 V, 50 Hz source, energy consumed by the heater = 2.3 units or 2.3 kWh in 1 hour Power consumed by the heater $=\frac{energy}{time \;eriod}=\frac{2.3kWh}{1 \; hour}$ $P_1=2.3 kW$ rms value of the input voltage $=V_{rms}=230V$ $P_1=\frac{V_{rms}^2}{R}$ $\Rightarrow \; 2.3 \times 10^3=\frac{230^2}{R}$ $\Rightarrow \; R=23\Omega$ (ii) When heater connected to 400 V (peak to peak) square wave source of 150Hz $V_{rms}$ value of the input voltage [/latex] $V_{rms}=\left[ \frac{1}{T}\int_{0}^{T}V^2 \; dt \right]^{1/2}$ $V_{rms}=\left[ \frac{1}{T}\left \{ \int_{0}^{T/2} 200^2 \; dt+ \int_{T/2}^{T} (-200)^2 \; dt\right \} \right]^{1/2}$ $V_{rms}=200V$ $P_2=\frac{V_{rms}^2}{R}=\frac{200^2}{23}\times 10^{-3}kW =1.739kW$

At f = 100Hz $|V_R|=|V_L|$ as R and L are series connected, current through R and L is same, so

R.M.S. value of d.c voltage $=V_{dc}^{(rms)}=3V$ R.M.S. value of a.c. voltage $=V_{ac}^{(rms)}=\left ( \frac{4}{\sqrt{2}} \right ) V$ Therefore, R.M.S. value of the voltage $\sqrt{3^{2+}\left ( \frac{4}{\sqrt{2}} \right )^2}$ $=\sqrt{9+8}=\sqrt{17}V$