Network Analysis
Network Basics
Practice questions from Network Basics.
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IncorrectThe I-V characteristics of the element between the nodes X and Y is best depicted by
Hence, Option (b) is correct
A nullator is defined as a circuit element where the voltage across the device and the current through the device are both zero. A series combination of a nullator and a resistor of value, R, will behave as a
V=0 and I=0
Overall circuit will also behave as a Nullator.
Hence, option (b) is correct.
The number of junctions in the circuit is
Junction node where 3 or more branches meet
Short circuit single junction
6 junctions (A to F)
For the circuit shown in the figure, and . The voltage in Volts is __________ (Round off to 1 decimal place).
Assume node-b as reference & applying nodal analysis.
8V – 48 = 0
V = 6 Volt
In the circuit shown below, the magnitude of voltage V1 in volts, across the 8 kΩ resistor is ________ . (round off to nearest integer)
The voltage across 8 kΩ is
V1 = 8 kI
Write KVL equation in first loop
75 = 2kI + 0.5V1
V1 = 100V
In the given circuit, for voltage Vy to be zero, value of β should be_________ (round off to 2 decimal places).
By KCL at node x,
4Vx – 24 + Vx + 2Vx – 2Vy = 0
7Vx – 2Vy = 24 ………..(i)
If Vy = 0
⇒ 7Vx = 24
By KCL at node y,
………..(ii)
The current I flowing in the circuit shown below in amperes (round off to one decimal place) is ____________.
Applying KVL
20-6= 10I
I= 1.4Amp
The current I flowing in the circuit shown below in amperes is ________.
By norton’s equivalent,
Since, I = 0A
The power supplied by the 25 V source in the figure shown below is ____________ W.
By KCL, current in
And
Power supplied by 25V source=
The equivalent resistance between the terminals A and B is __________ .
The circuit can be redrawn as
In the circuit shown below, the maximum power transferred to the resistor R is _____W.
For maximum power transfer to R, we need to determine Thevenin equivalent across R.
By KVL
For
Short circuit the voltage sources
In the circuit shown below, the value of capacitor C required for maximum power to be transferred to the load is
1 nF
1 mF
10 mF
For maximum power transfer ZL = Zs*
Since
For ZL to be purely real
5000 c2 – 100c + 0.5 = 0
For the network given in figure below, the Thevenin’s voltage Vab is
Apply Source Transformation,
Apply KVL in first loop
Apply KVL in second loop
Solving (i) and (ii) I1 = 1.9A, I2 = 1.75A
and are the input resistances of circuits as shown below. The circuit extends infinitely in the direction shown. Which one of the following statements is TRUE?
The given circuits are,
On comparing the two circuits we can obtain second circuit from first one as shown below,
In the portion of a circuit shown, if the heat generated in resistance is 10 calories per second, then heat generated by the resistance, in calories per second, is__________.
Current through resistance
Current through resistor
Heat generated in resistance,
Heat generated in resistance
Substituting the value of I2
Heat generated in resistance
In the given circuit, the current supplied by the battery, in ampere, is______________
The given system is shown below,
By KCL:
………..…(1)
Now applying KVL in outer loop,
..……….…(2)
Solving (1) & (2),
Current supplied by battery
In the circuit shown below, the node voltage is ________________V.
Let voltage of node A be V volts.
Applying KCL at node A,
……….…(1)
Also, ..……..…(2)
Substituting (2) in (1),
The voltages developed across the and 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?
The directions of currents have been marked in the figure below,
Current through resistor =
Current through resistor
KCL at N2,
Power delivered by 5V source = 5 x 1 = 5W
In the given circuit, the parameter k is positive, and the power dissipated in the 2 resistor is 12.5W. The value of k is ____________.
Current through
By KCL, ……………(1)
From (1),
The three circuit elements shown in the figure are part of an electric circuit. The total power absorbed by the three circuit elements in watt is _______________.
For the 100V source, current goes into positive terminal
So, power absorbed= 100 × 10 = 1000 W
For the 80 V source, current comes out of positive terminal
Power absorbed = –(80 × 8) = – 640 W (power delivered)
Current in 15V source = 10 – 8 = 2A (downwards)
Power absorbed = –(15 × 2) = –30 W (power delivered)
Total power absorbed = 1000 – 640 – 30 = 330 W
An incandescent lamp is marked 40W, 240V. If resistance at room temperature is and temperature coefficient of resistance is , then its ‘ON’ state filament temperature is is approximately______________.
Resistance in ON state =
1440 = 120
= = 2444.44
T – Troom = 2444.44
T = 2444.44 + 26 = 2470.44
In the figure, the value of resistor R is (25 + I/2) ohms, where I is the current in amperes. The current I is ________________.
By KVL, I =
Since, R = (25 + I/2) ohms
I =
I = 10 A
Current cannot be negative as resistance cannot deliver current.
The power delivered by the current source, in the figure is_________.
Assuming the ground node as shown below,
Applying KCL at ‘V’,
V = 1.5 V
Power supplied by current source = VI =1.5 × 2= 3W
Consider a delta connection of resistors and its equivalent star connection as shown below. If all elements of the delta connection are scaled by a factor k, k>0, the element of the corresponding star equivalent will be scaled by a factor of
k
1/k
After scaling
Similarly,
Hence, all resistors of star connected system are scaled by a factor of k.
If , then is
The given system is,
Since current in AB & CD branches are independent of networks on both terminals, so considering both the networks as independent network as shown below.
Since
Current in 1 A resistor (from D to C)
If the resistor draws a current of 1 A as shown in the figure, the value of resistance R is
Applying KCL at node A
Now applying KVL in loop (2)
The current through the 2kΩ resistance in the circuit shown is
Since the bridge is balanced
Node C and node D will be at same potential. So the current in resistor will be zero
How many 200W/220V incandescent lamps connected in series would consume the same total power as a single 100W/220V incandescent lamp?
For a single 200W/220V lamp,
For a single 100W/220V lamp
So, second case acts as series combination of two lamps from first case.
Hence if two 200W/220V bulbs are connected in series then they will consume same power as a single 100W/220V bulb
Assuming ideal elements in the circuit shown below, the voltage will be
By KCL, i=1A
Applying KVL in circuit
In the circuit shown in the figure. The value of the current i will be given by
By Potential Division,
And by ohm’s law
Volts
Applying KVL in 2nd loop
From equation (1)
The current i(t) sketched in the figure flows through an initially uncharged 0.3nF capacitor.
The charge stored in the capacitor at t = 5µs, will be
Charge stored at is equal to area of i vs t curve up to
The capacitor charged up to 5µs, as per the current profile given in the figure, is connected across an
inductor of 0.6mH. Then the value of voltage across the capacitor after 1µs will approximately be
Since,
Voltage across the capacitor at this instant [before connecting inductor]
When inductor is connected, circuit will have inductor & capacitor only. So sinusoidal current & voltage will be produced due to LC oscillations.
Note: This can be derived using Laplace Transform else remember this expression whenever capacitor is initially charged to a certain voltage.
Frequency,
At
A 3 V dc supply with an internal resistance of 2Ω supplies a passive non-linear resistance characterized by the relation. The power dissipated in the non linear resistance is
By KVL
Since, characteristics of non-linear resistor is
Since, non-linear is passive. So it cannot act as source, so is rejected
Hence,
Power dissipated in non-linear resistance
Figure the value of R is
Combining the two resistances 10Ω in parallel
Using KVL in loop
In figure, the value of the source voltage is
Applying KCL at node A
Applying KVL in loop (2)
In figure, Ra, Rb and Rc are 20Ω, 10Ω and 10Ω respectively. The resistance R1, R2 and R3 in Ω of an equivalent star-connection are
In figure, the value of resistance R in Ω is
In 2nd loop
Apply KVL in 1st loop
Applying KVL in 2nd loop
Figure shows the waveform of the current passing through an inductor of resistance 1Ω and inductance 2 H. The energy absorbed by the inductor in the first four seconds is
By analyzing current time graph
Energy stored in inductance
This could have directly been found by,
Energy dissipated in resistance
Total energy absorbed by inductor
In Figure, the potential difference between points P and Q is
Applying KCL at point ‘P’
Applying KCL at point ‘Q’
Potential difference between points P and Q
In Figure the value of R is
Applying KCL at node ‘A’
Applying KCL at node ‘C’
From equation (1)
From equation (1)
In the resistor network shown in Figure, all resistor values are 1Ω. A current of 1A passes from terminal a to terminal b, as shown in the figure. Calculate the voltage between terminals a and b. [Hint: You may exploit the symmetry of the circuit].
Since all resistor are equal, so both Wheatstone bridges are balanced and both the diagonal resistors are neglected as shown below,
Applying delta to star conversion in upper triangle
The circuit can then further be simplified as shown below,
So,
This method is wrong as whitestone bridge Is not balance because we must used equivalent resistance while checking balancing.
Correct method
Given circuit can be redrawn as
Covert Delta (a12) to star & delta (b34) to star
Again convert Delta to star
Two incandescent light bulbs of 40W and 60W rating are connected in series across the mains.
Then
The individual light bulbs and their connection is shown below,
Since the first bulb consumes 40W when connected to a 220V supply.
Similarly, the second bulb consumes 60W when connected to a 220V supply.
Power consumption in series combination
= 24W
So total consumption = 24W
Current in series combination
Power consumption in 40W bulb = = 14.4W
Power consumption in 60W bulb = = 9.6W
Hence power consumption in 40W bulb is more, so 40W bulb glows brighter.
Consider the star network shown in Figure. The resistance between terminals A and B with C open is 6Ω, between terminals B and C with A open is 11Ω, and between terminals C and A with B open is 9Ω. Then
When terminal ‘C’ is open, then resistance between
‘A’ and ‘B’
----------------(1)
When terminal ‘A’ is open then resistance between ‘B’ & ‘C’
----------------(2)
Similarly when terminal ‘B’ is open then resistance between ‘A’ & ‘C’
----------------(3)
By solving equation 1, 2 & 3
, ,
An ideal voltage source will charge an ideal capacity
Ideal voltage source has zero internal resistance; R = 0.
Time constant
Hence capacitor will charge instantaneously.
A practical current source is usually represented by
A practical current source is usually represented by a resistance in parallel with an ideal current source.
It can also be derived from a practical voltage source which is represented as a voltage source in series with internal resistance by Source Transformation Technique.
Energy stored in a capacitor over a cycle, when excited by an a.c. source is
When a capacitor is excited by an AC source, the capacitor stores energy for half time period and delivers that energy in another time period. Hence total energy stored in a capacitor over a complete cycle, when excited by an AC source is zero.
This can be derived by plotting current and voltage and as current leads the voltage the product of current and voltage is positive for half cycle i.e. capacitor stores energy in half cycle and product is negative in other half cycle so capacitor delivers energy.
This can also be explained as, the final and initial value of voltage will be same after a complete cycle and since energy stored in a capacitor is . So, initial and final values of stored energy will also be same. Hence, energy stored will be zero.
For the two port network shown in figure, the admittance matrix is . Then Find
\(Y_{11} + Y_{12} + Y_{22} + Y_{21} = ?\)
Applying KCL at node ‘A’
-------------(1)
Applying KCL at node ‘B’
-------------- (2)
From equation 1 & 2
\(Y_{11} + Y_{12} + Y_{22} + Y_{21} = 0.2 - 0.1 - 0.1 + 0.2 = 0.2\)
A 10V battery with an internal resistance of 1Ω is connected across a nonlinear load whose V-I characteristic is given by . The current delivered by the battery is __________ A
For non linear resistor ----------------(1)
Applying KVL
---------------(2)
From equation 1 & 2
V = 5, -14 Volts
Taking Positive sign, as the non linear resistor is behaving as a load, so current direction should be from higher voltage to lower voltage which agrees with the direction shown.
V = -14 is neglected as current direction will remain same in that case but the resistor will start delivering power.
Hence,
From equation 2
The values of E and I for the circuit shown in figure, are_________V and _________A. Then E+I= ?
Assuming different currents in the circuit branches the circuit is shown below,
Voltage across resistor
By KCL,
by KVL,
Current through resistor = 10 + 3
By KVL,
Then E+I=31+13=44
The voltage and current waveforms for an element are shown in figure. The circuit element is _________And its value is
The current and voltage waveforms are shown below,
For
and
So,
Comparing with
So the element is an Inductor and its value is 2H.
In the circuit shown in figure X is an element which always absorbs power. During a particular operations, it sets up a current of 1 amp in the direction shown and absorbs a power . It is possible that X can absorb the same power for another current i, the value of this current is
None of these
By Energy Conservation, total power absorbed by resistor and element ‘X’ is equal to power supplied by voltage source.
at I = 1 amp
In second case,
let current is I
So, at I = 5 Amp, ‘X’ will absorb same power of 5W.
All the resistances in figure are each. The value of current ‘I’ is
Each resistances is given as 1Ω
Two resistances are connected in parallel,
The above connection can be simplified as shown below,
The resistances can then be combined in series as
Then two branches can be combined in parallel
The circuit then looks like as shown below,
Again, the series combination yields,
Then branches can be combined in parallel as,
The simplified circuit is shown below,
In series,
