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Find the node voltage VA.<br> <img src="/images/question-image/electrical-engineering/branch,-loop-and-node-analyses/1526016899-2.png" title="Branch, Loop and Node Analyses mcq question image" alt="Branch, Loop and Node Analyses mcq question image">
A
6 V
B
12 V
C
4.25 V
D
3 V
Correct Answer:
4.25 V
Find the node voltage VA.
A
518 mV
B
5.18 V
C
9.56 V
D
956 mV
What is the voltage drop across R1?
A
850 mV
B
7.82 V
C
9.18 V
D
918 mV
Find branch current IR2.
A
5.4 mA
B
-5.4 mA
C
113.0 mA
D
119.6 mA
Using the mesh current method, find the branch current, IR1, in the above figure.
A
115 mA
B
12.5 mA
C
12.5 A
D
135 mA
What is the current through R2?
A
3.19 A
B
319 mA
C
1.73 A
D
173 mA
Consider a conducting loop of radius a and total loop resistance R placed in a region with a magnetic field B thereby enclosing a flux $${\phi _0}$$ . The loop is connected to an electronic circuit as shown, the capacitor being initially uncharged.
If the loop is pulled out of the region of the magnetic field at a constant speed u, the final output voltage V
out
is independent of
A
$${\phi _0}$$
B
u
C
R
D
C
An infinitely long wire carrying a current $$I\left( t \right) = {I_0}\cos \left( {\omega t} \right)$$ is placed at a distance a from a square loop of side a as shown in the figure. If the resistance of the loop is R, then the amplitude of the induced current in the loop is
A
$$\frac{{{\mu _0}}}{{2\pi }} \cdot \frac{{a{I_0}\omega }}{R}\ln 2$$
B
$$\frac{{{\mu _0}}}{\pi } \cdot \frac{{a{I_0}\omega }}{R}\ln 2$$
C
$$\frac{{2{\mu _0}}}{\pi } \cdot \frac{{a{I_0}\omega }}{R}\ln 2$$
D
$$\frac{{{\mu _0}}}{{2\pi }} \cdot \frac{{a{I_0}\omega }}{R}$$
A sinusoidal input voltage V
in
of frequency ω is fed to the circuit shown in the figure, where C
1
≫ C
2
. If V
m
is the peak value of the input voltage, then output voltage (V
out
) is
A
$$2{V_m}$$
B
$$2{V_o}\sin \omega t$$
C
$$\sqrt 2 {V_m}$$
D
$$\frac{{{V_m}}}{2}\sin \omega t$$
Let $${I_1}$$ and $${I_2}$$ represents mesh currents in the loop abcda and befcb respectively. The correct expression describing Kirchhoff's voltage loop law in one of the following loops is,
A
$$30{I_1} - 15{I_2} = 10$$
B
$$ - 15{I_1} + 20{I_2} = - 20$$
C
$$30{I_1} - 15{I_2} = - 10$$
D
$$ - 15{I_1} + 20{I_2} = 20$$
Assuming an ideal voltage source, Thevenin's resistance and Thevenin's voltage respectively for the above circuit are
A
15 $$\Omega $$ and 7.5 V
B
20 $$\Omega $$ and 5 V
C
10 $$\Omega $$ and 10 V
D
30 $$\Omega $$ and 15 V
