# Electric Circuits

 Question 1
A benchtop dc power supply acts as an ideal 4 A current source as long as its terminal voltage is below 10 V. Beyond this point, it begins to behave as an ideal 10 V voltage source for all load currents going down to 0 A. When connected to an ideal rheostat, find the load resistance value at which maximum power is transferred, and the corresponding load voltage and current.
 A Short, $\infty$ A, 10 V B Open, 4 A, 0 V C 2.5 $\Omega$, 4 A, 10 V D 2.5 $\Omega$, 4 A, 5 V
GATE EE 2020      Network Theorems
Question 1 Explanation:

Maximum power transistor of VI product is maximum. If draw the the curve, it intersect (10, 4) that will give maximum power. The terminal voltage is 10 V (Load voltage) and current is 4 A (Load current). Load resistance is $\frac{10}{4}=2.5\Omega$ .
 Question 2
The Thevenin equivalent voltage, $V_{TH}$, in V (rounded off to 2 decimal places) of the network shown below, is _______ .
 A 5 B 7 C 14 D 6
GATE EE 2020      Network Theorems
Question 2 Explanation:
Only voltage source 4V is there and current source 5A is open circuited

From the above circuit,
$V_{TH1}=4V$

Case-II:
Only current source 5A is there and voltage source 4V is short circuited.

From the above circuit,
$V_{TH2}=2 \times 5=10V$
By applying superposition theorem,
$V_{TH}=V_{TH1}+V_{TH2}=10+4=14V$
 Question 3
$x_R\; and \; x_A$ are, respectively, the rms and average values of $x(t) = x(t - T)$, and similarly, $y_R\; and \; y_A$ are, respectively, the rms and average values of $y(t) = kx(t). k, \;T$ are independent of t. Which of the following is true?
 A $y_A=kx_A;\; y_R=kx_R$ B $y_A=kx_A;\; y_R\neq kx_R$ C $y_A \neq kx_A;\; y_R= kx_R$ D $y_A \neq kx_A;\; y_R\neq kx_R$
GATE EE 2020      Network Theorems
Question 3 Explanation:
Given that,
\begin{aligned} x(t)&=x(t-T) \; \text{ i.e. periodic signal}\\ \text{Average of } x(t)&=x_{A}\\ \text{Rms of }x(t)&=x_{R} \\ \text{Average of }y(t)&=y_{A}\\ \text{Rms of } y(t)&=y_{R} \\ y(t)&=k_{x}(t_{0}) \; \; ....(i) \\ &\text{Using equation(i),}\\ \text{Average of }y(t)&=k \times \text{ Average of }x(t) \\ y_{A}&=kx_{A}\\ \text{Power of }y(t)&=|k|^{2}\text{ Power of }x(t) \\ Rms^{2} \text{ of } y(t)&=|k|^{2}\; Rms^{2}\text{ of }x(t) \\ y_{R}^{2}=|k|^{2}\cdot x_{R}^{2} \\ y_{R}=|k|x_{R}\end{aligned}
 Question 4
The current flowing in the circuit shown below in amperes is _____
 A 0 B 1 C 2 D 4
GATE EE 2019      Network Theorems
Question 4 Explanation:

By Millman'e theorem,
$E=\frac{\frac{200}{50}+\frac{160}{40}-\frac{100}{25}-\frac{80}{20}}{\frac{1}{50}+\frac{1}{40}+\frac{1}{25}+\frac{1}{20}}=0V$
$\frac{1}{R}=\frac{1}{50}+\frac{1}{40}+\frac{1}{25}+\frac{1}{20}$
Simplified circuit,
$\therefore \;\;I=0A$
 Question 5
A 0.1$\mu F$ capacitor charged to 100 V is discharged through a 1 $k\Omega$ resistor. The time in ms (round off to two decimal places) required for the voltage across the capacitor to drop to 1V is ______
 A 0.25 B 0.65 C 0.45 D 0.85
GATE EE 2019      Steady state AC Analysis
Question 5 Explanation:

\begin{aligned} v_c(t)&=V_0e^{-t/\tau } \\ V_0&=100V \\ \tau &=RC=(10^3)(10^{-7}) \\ &=10^{-4}sec \\ \therefore \;v_c(t)&=100e^{-10^4 t }V \end{aligned}
Let the time required by the voltage across the capacitor to drop to 1 V is $t_1,$
\begin{aligned} \therefore \; v_c(t_1)&=100e^{-10^4t_1} \\ \text{But, } v_c(t_1)&=0 \\ \text{So, }0&=100e^{-10^4t_1} \\ t_1&=0.46msec \end{aligned}
 Question 6
The line currents of a three-phase four wire system are square waves with amplitude of 100 A. These three currents are phase shifted by 120$^{\circ}$ with respect to each other. The rms value of neutral current is
 A 0 A B $\frac{100}{\sqrt{3}}$ A C 100 A D 300 A
GATE EE 2019      Magnetically Coupled Circuits, Network Topology and Filters
Question 6 Explanation:

$I_N=I_a+I_b+I_c$
$(I_N)_{rms}=100A$
 Question 7
The current I flowing in the circuit shown below in amperes (round off to one decimal place) is ____
 A 0.8 B 1.1 C 1.4 D 1.9
GATE EE 2019      Basics
Question 7 Explanation:

Applying nodal at node x,
$-I-2+\frac{V_x-5I}{3}=0$
$-3I-6+V_x-5I=0$
$\Rightarrow \; 8I=V_x-6\;\;...(i)$
As, $I=\frac{20-V_x}{2}$
$\Rightarrow \; V_x=20-2I\;\;...(ii)$
Substituting (ii) in (i),
$8I=20-2I-6$
$10I=14$
$I=1.4A$
 Question 8
The voltage v(t) across the terminals a and b as shown in the figure, is a sinusoidal voltage having a frequency $\omega$=100 radian/s. When the inductor current i(t) is in phase with the voltage v(t), the magnitude of the impedance Z (in $\Omega$) seen between the terminals a and b is ________ (up to 2 decimal places).
 A 25 B 50 C 100 D 150
GATE EE 2018      Resonance and Locus Diagrams
Question 8 Explanation:
At resonance imaginary part of $Z_{eq}=0$
Real of $Z_{eq}=\frac{R_1 X_c^2}{R_1^2+X-c^2}$
$\;\;=\frac{100 \times 100 \times 100}{100^2+100^2}=50\Omega$
 Question 9
The voltage across the circuit in the figure, and the current through it, are given by the following expressions:
$v(t) = 5 - 10 cos(\omega t + 60^{\circ}) V$
$i(t) = 5 + X cos(\omega t) A$
where $\omega =100 \pi$ radian/s. If the average power delivered to the circuit is zero, then the value of X (in Ampere) is _____ (up to 2 decimal places).
 A 5 B 10 C 15 D 20
GATE EE 2018      Steady State AC Analysis
Question 9 Explanation:
Given that,
$v(t)=5-10 \cos (\omega t+60^{\circ})$
$i(t)=5+C \cos (\omega t-0^{\circ})$
$P_{req}=0$
$0=5 \times 5 +\frac{1}{2}[(-10)(X)\cos(60^{\circ})]$
$-25=\frac{1}{2}[(-10)(X)\cos(60^{\circ})]$
$X=10$
 Question 10
The equivalent impedance $Z_{eq}$ for the infinite ladder circuit shown in the figure is
 A j12 $\Omega$ B -j12 $\Omega$ C j13 $\Omega$ D 13 $\Omega$
GATE EE 2018      Basics
Question 10 Explanation:

$Z_1=j9$
$Z_2=j5-j1=j4$
$Z_{eq}=Z_1+\frac{Z_2Z_{eq}}{Z_2+Z_{eq}}$
By solving above equation,
$Z_{eq}=j12$

There are 10 questions to complete.