Questions & Answers of KCL, KVL, Node and Mesh Analysis

•  (A)  R_A=R_B
•  (B)  R_A=R_B=0
•  (C)  R_A< R_B
•  (D)  R_B= R_A /(1+R_A)

In the given circuit, the current supplied by the battery, in ampere, is _______.

In the circuit shown below, the node voltage V_A is ___________ V.

$v\left(t\right)=100\sin \left(\omega t\right)$ ,

$i\left(t\right)=10\sin \left(\omega t-{60}^{°}\right)+2\sin \left(3\omega t\right)+5\sin \left(5\omega t\right)$

The average power consumed by the load, in W, is___________.

•  (A)  0
•  (B)  5
•  (C)  10
•  (D)  20

The voltages developed across the 3 Ω and 2 Ω resistors shown in the figure are 6V and 2V respectively, with the polarity as marked. What is the power (in Watt) delivered by the 5V voltage source?

•  (A) 5
•  (B) 7
•  (C) 10
•  (D) 14

In the given circuit, the parameter k is positive, and the power dissipated in the 2 $\mathrm{\Omega }$ resistor is 12.5W. The value of k is________.

The line A to neutral voltage is A10∠15^oV for a balanced three phase star-connected load with phase sequence ABC. The voltage of line B with respect to line C is given by

•  (A) $10\sqrt{3}\angle 105°V$
•  (B) $10\angle 105°V$
•  (C) $10\sqrt{3}\angle -75°V$
•  (D) $-10\sqrt{3}\angle 90°V$

The power delivered by the current source, in the figure, is ________.

In the circuit shown below, the current through the inductor is

•  (A) $\frac{2}{1+j}A$
•  (B) $\frac{-1}{1+j}A$
•  (C) $\frac{1}{1+j}A$
•  (D) 0

A two-phase load draws the following phase currents: $i_1\left(t\right)=I_m\sin \left(\omega t-{\Phi }_1\right)$, $i_2\left(t\right)=I_m\cos \left(\omega t-{\Phi }_2\right)$.These currents are balanced if ${\Phi }_1$ is equal to

•  (A) $-{\Phi }_2$
•  (B) ${\Phi }_2$
•  (C) $\left(\mathrm{\pi }/2-{\Phi }_2\right)$
•  (D) $\left(\mathrm{\pi }/2+{\Phi }_2\right)$

The average power delivered to an impedance (4-j3)$\Omega$ by a current $5\cos \left(100\pi t+100\right)A$ is

•  (A) 44.2 W
•  (B) 50 W
•  (C) 62.5 W
•  (D) 125 W

If V_A-V_B=6V, then V_C-V_D is

•  (A) –5 V
•  (B) 2 V
•  (C) 3 V
•  (D) 6 V

The r.m.s value of the current  i (t) in the circuit shown below is 

•  (A) $\frac{1}{2}A$
•  (B) $\frac{1}{\sqrt{2}}A$
•  (C) $1A$
•  (D) $\sqrt{2}A$

The voltage applied to a circuit is 100 $\sqrt{2}$ cos (100$\pi$t) volts and the circuit draws a current of 10 $\sqrt{2}$ sin (100$\pi$t + $\pi$ / 4) amperes. Taking the voltage as the reference phasor, the phasor representation of the current in amperes is

•  (A) $10\sqrt{2}\angle -\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$
•  (B) $10\angle -\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$
•  (C) $10\angle +\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$
•  (D) $10\sqrt{2}\angle +\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

The input voltage given to a converter is
$v_i=100\sqrt{2}\sin \left(100\pi t\right)\mathrm{V}$

The current drawn by the converter is
$i_i=\left(10\sqrt{2}\sin \left(100\pi t-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)+5\sqrt{2}\sin \left(300\pi t+\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)+2\sqrt{2}\sin \left(500\pi t-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)\right) \mathrm{A}$

The input power factor of the converter is

•  (A) 0.31
•  (B) 0.44
•  (C) 0.5
•  (D) 0.71

The input voltage given to a converter is
$v_i=100\sqrt{2}\sin \left(100\pi t\right)\mathrm{V}$

The current drawn by the converter is
$i_i=\left(10\sqrt{2}\sin \left(100\pi t-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)+5\sqrt{2}\sin \left(300\pi t+\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)+2\sqrt{2}\sin \left(500\pi t-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)\right) \mathrm{A}$

The active power drawn by the converter is

•  (A) 181 W
•  (B) 500 W
•  (C) 707 W
•  (D) 887 W

An RLC circuit with relevant data is given below.

The power dissipated in the resistor R is

•  (A) 0.5W
•  (B) 1W
•  (C) $\sqrt{2}$W
•  (D) 2W

An RLC circuit with relevant data is given below.

The current ${\underline{I}}_C$ in the figure above is

•  (A) −j2 A
•  (B) -j$\frac{1}{\sqrt{2}}$A
•  (C) +j$\frac{1}{\sqrt{2}}$A
•  (D) +j2A

As shown in the figure, a 1Ω resistance is connected across a source that has a load line v+ i = 100. The current through the resistance is

•  (A) 25A
•  (B) 50A
•  (C) 100A
•  (D) 200A

If the 12Ω resistor draws a current of 1A as shown in the figure, the value of resistance R is

•  (A) 4Ω
•  (B) 6Ω
•  (C) 8Ω
•  (D) 18Ω

The current through the 2 kΩ resistance in the circuit shown is

•  (A) 0 mA
•  (B) 1 mA
•  (C) 2 mA
•  (D) 6 mA

The equivalent capacitance of the input loop of the circuit shown is

•  (A) 2 µF
•  (B) 100 µF
•  (C) 200 µF
•  (D) 4 µF

For the circuit shown, find out the current flowing through the 2Ω resistance. Also identify the changes to be made to double the current through the 2Ω resistance

•  (A) (5A ; Put V_s = 20V)
•  (B) (2A ; Put V_s = 8V)
•  (C) (5A ; Put I_s = 10A)
•  (D) (7A ; Put I_s = 12A)

The Thevenin’s equivalent of a circuit operation at ω = 5 rads/s, has $V_{\mathrm{oc}}=3.71\angle -15.9°V\mathrm{and} Z_{\mathrm{o}}=2.38-j0.667\Omega .$At this frequency, the minimal realization of the Thevenin’s impedance will have a

•  (A) resistor and a capacitor and an inductor
•  (B) resistor and a capacitor
•  (C) resistor and an inductor
•  (D) capacitor and an inductor

Assuming ideal elements in the circuit shown below, the voltage V_ab will be

•  (A) -3V
•  (B) 0 V
•  (C) 3 V
•  (D) 5 V

In the circuit shown in the figure, the value of the current i will be given by

•  (A) 0.31 A
•  (B) 1.25 A
•  (C) 1.75 A
•  (D) 2.5 A

The state equation for the current I₁ in the network shown below in terms of the voltage V_x and the independent source V, is given by

•  (A) $\frac{d{I}_1}{dt}=-1.4V_x-3.75I_1+\frac{5}{4}V$
•  (B) $\frac{d{I}_1}{dt}=1.4V_x-3.75I_1-\frac{5}{4}V$
•  (C) $\frac{d{I}_1}{dt}=-1.4V_x+3.75I_1+\frac{5}{4}V$
•  (D) $\frac{d{I}_1}{dt}=-1.4V_x+3.75I_1-\frac{5}{4}V$

In the figure given below all phasors are with reference to the potential at point "O". The locus of voltage phasor V_YX as R is varied from zero to infinity is shown by

A 3 V DC supply with an internal resistance of 2 Ω supplies a passive non-linear resistance characterized by the relation V_NL = I²_NL. The power dissipated in the non-linear resistance is

•  (A) 1.0 W
•  (B) 1.5 W
•  (C) 2.5 W
•  (D) 3.0 W