Question 1 |
The current I in the given network is
0 A | |
2.38 \angle -96.37^{\circ} A | |
2.38 \angle 143.63^{\circ} A | |
2.38 \angle -23.63^{\circ} A |
Question 1 Explanation:
I=-[I_{1}+I_{2}]
I=-\left [ \frac{120\angle -90^{\circ}}{80-j35}+\frac{120\angle -30^{\circ}}{80-j35} \right ]
I=2.38\angle 143.7^{\circ}
Question 2 |
Consider the network shown below with R_{1} = 1\Omega , R_{2} = 2\Omega \; and \; R_{3} = 3\Omega . The network is connected to a constant voltage source of 11V.
The magnitude of the current (in amperes, accurate to two decimal places) through the source is _______.
The magnitude of the current (in amperes, accurate to two decimal places) through the source is _______.
5 | |
8.0 | |
8.7 | |
9.5 |
Question 2 Explanation:
As the network is symmetric,
V_{A}= V_{B} and V_{C}= V_{D}
So current throught R_{2} resistor is zero and V_{A}=V_{B} and V_{C}=V_{D},electrically the circuit can reduced as,
Total resistance,
\begin{aligned} R_{T} &=2\left(R_{1} \| R_{1}\right)+\left(R_{1}\left\|R_{1}\right\| R_{3} \| R_{3}\right) \\ &=R_{1}+\left(\frac{R_{1}}{2} \| \frac{R_{3}}{2}\right) \end{aligned}
Given that, R_{1}=1 \Omega and R_{3}=3 \Omega
\begin{aligned} \text{So,}\qquad R_{T} &=1+\left(\frac{1}{2} \| \frac{3}{2}\right) \Omega=1+\frac{3 / 2}{4}=\frac{11}{8} \Omega \\ I &=\frac{11 \mathrm{V}}{R_{T}}=\frac{11}{(11 / 8)}=8 \mathrm{A} \end{aligned}
Question 3 |
A connection is made consisting of resistance A in series with a parallel combination of
resistances B and C. Three resistors of value 10\Omega, 5\Omega, 2\Omega are provided. Consider all possible permutations of the given resistors into the positions A, B, C, and identify the configurations with maximum possible overall resistance, and also the ones with minimum possible overall resistance. The ratio of maximum to minimum values of the resistances (up to second decimal place) is ____________.
0.466 | |
2.14 | |
16.758 | |
6.098 |
Question 3 Explanation:
The connection of resistors is as shown below:
Given resistor values are : 10 \Omega, 5 \Omega, 2 \Omega
The maximum resistance possible is,
\begin{aligned} R_{T(\max )} &=10 \Omega+(5 \Omega \| 2 \Omega) \\ &=\left(10+\frac{10}{7}\right) \Omega=\frac{80}{7} \Omega \end{aligned}
The minimum resistance possible is,
\begin{aligned} R_{T(\min )} &=2 \Omega+(10 \Omega \| 5 \Omega) \\ &=\left(2+\frac{10}{3}\right) \Omega=\frac{16}{3} \Omega \\ \frac{R_{T(\max )}}{R_{T(\min )}} &=\frac{80 / 7}{16 / 3}=\frac{15}{7}=2.143 \end{aligned}
Given resistor values are : 10 \Omega, 5 \Omega, 2 \Omega
The maximum resistance possible is,
\begin{aligned} R_{T(\max )} &=10 \Omega+(5 \Omega \| 2 \Omega) \\ &=\left(10+\frac{10}{7}\right) \Omega=\frac{80}{7} \Omega \end{aligned}
The minimum resistance possible is,
\begin{aligned} R_{T(\min )} &=2 \Omega+(10 \Omega \| 5 \Omega) \\ &=\left(2+\frac{10}{3}\right) \Omega=\frac{16}{3} \Omega \\ \frac{R_{T(\max )}}{R_{T(\min )}} &=\frac{80 / 7}{16 / 3}=\frac{15}{7}=2.143 \end{aligned}
Question 4 |
In the figure shown, the current i (in ampere) is __________
0 | |
-1 | |
-2 | |
-4 |
Question 4 Explanation:
Using KCL at V_{1}
\frac{V_{1}}{1}+\frac{V_{1}-8}{1}+\frac{V_{1}-8}{1}+\frac{V_{1}}{1}=0
or V_{1}=4 V
considering KVL, we get
Question 5 |
In the circuit shown in the figure, the magnitude of the current (in amperes) through R_{2} is ___
2 | |
3 | |
4 | |
5 |
Question 5 Explanation:
Using KVL in the outer loop
60-5\left(0.16 V_{x}\right)-\frac{V_{x}}{5} \times 3-V_{x}=0
\text{or}\quad v_{x}=25 \mathrm{V}
\therefore The current flowing through
R_{2}=\frac{V_{x}}{5}=\frac{25}{5}=5 \mathrm{A}
Question 6 |
In the given circuit, each resistor has a value equal to 1 \Omega.
What is the equivalent resistance across the terminals a and b?
What is the equivalent resistance across the terminals a and b?
1/6 \Omega | |
1/3 \Omega | |
9/20 \Omega | |
8/15 \Omega |
Question 6 Explanation:
R_{eq}\Rightarrow
By using delta to star conversion
Again by star to delta conversion
\begin{aligned} R_{\mathrm{ab}} &=\{4 \| 1)+(4 \| 1)\}\|\{1 \| 4)\} \\ &=\left(\frac{4}{5}+\frac{4}{5}\right) \| \frac{4}{5}=\frac{8}{15} \Omega \end{aligned}
By using delta to star conversion
Again by star to delta conversion
\begin{aligned} R_{\mathrm{ab}} &=\{4 \| 1)+(4 \| 1)\}\|\{1 \| 4)\} \\ &=\left(\frac{4}{5}+\frac{4}{5}\right) \| \frac{4}{5}=\frac{8}{15} \Omega \end{aligned}
Question 7 |
An AC voltage source V = 10 sin(t) volts is applied to the following network. Assume that R_{1}=3k \Omega , R_{2}=6k \Omega and R_{3}=9k \Omega , and that the diode is ideal.
RMS current I_{rms} (in mA) through the diode is ________
RMS current I_{rms} (in mA) through the diode is ________
0.5 | |
1 | |
1.5 | |
2 |
Question 7 Explanation:
The equivalent resistance across terminal ah ( outer loop) is
\begin{aligned} V &=\frac{I}{3} \times 3 \mathrm{k} \Omega+\frac{I}{6} \times 6 \mathrm{k} \Omega+\frac{I}{3} \times 9 \mathrm{k} \Omega \\ V &=5 I \\ \text{or}\quad \frac{V}{I} &=5 \mathrm{k} \Omega \end{aligned}
For half wave rectifier
\begin{aligned} I_{\mathrm{rms}} &=\frac{I_{m}}{(2)}=\frac{10 \sin t}{5 \mathrm{k} \Omega}=2 \sin t \mathrm{m} \mathrm{A} \\ \therefore \quad I_{\mathrm{rms}} &=\frac{I_{m}}{2}=1 \mathrm{m} \mathrm{A} \end{aligned}
Question 8 |
In the circuit shown, the voltage V_{X} (in Volts) is _______.
7 | |
8 | |
9 | |
10 |
Question 8 Explanation:
\begin{aligned} \frac{V_{x}}{20}+\frac{V_{x}-V_{y}}{10}+0.5 V_{x}&=5 \mathrm{A} \\ V_{x}\left[\frac{1}{20}+\frac{1}{10}+0.5\right]&=5+\frac{V_{y}}{10} \\ 13 \mathrm{V}_{x}&=100+2 \mathrm{V}_{y} \quad\ldots(i)\\ \text { and also, } V_{y}&=0.25 \mathrm{V}_{x} \quad\dots(ii) \end{aligned}
Solving equations (i) and (ii), we have
\begin{aligned} 52 V_{y} &=100+2 V_{y} \\ 50 V_{y} &=100 \Rightarrow V_{y}=2 \mathrm{V} \\ V_{x} &=4 V_{y}=8 \mathrm{V} \end{aligned}
Question 9 |
In the given circuit, the values of V_{1} \; and \; V_{2} respectively are
5 V, 25 V | |
10 V,30 V | |
15 V,30 V | |
0 V , 20V |
Question 9 Explanation:
Current flowing through both the parallel 4 \Omega will
be I.
\text { So, } V_{2}=4(I+I+2 I)+4 I \quad \text { by KVL }
I+I+2 I=5 \quad \text{by KCL}
\begin{aligned} I &=\frac{5}{4} A \\ V_{2} &=4 \times 5+\frac{4 \times 5}{4}=25 \mathrm{V} \\ V_{1} &=4 I=\frac{4 \times 5}{4}=5 \mathrm{V} \end{aligned}
Question 10 |
In the network shown in the figure, all resistors are identical with R = 300 \Omega. The resistance R_{ab} (in \Omega) of the network is ______.
95 | |
100 | |
110 | |
120 |
Question 10 Explanation:
Modifying the given circuit
\begin{aligned} R_{a b} &=\left(\frac{1}{2 R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{2 R}\right)^{-1} \\ &=\frac{R}{3}=\frac{300}{3}=100 \Omega \end{aligned}
\begin{aligned} R_{a b} &=\left(\frac{1}{2 R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{2 R}\right)^{-1} \\ &=\frac{R}{3}=\frac{300}{3}=100 \Omega \end{aligned}
There are 10 questions to complete.
