ECE - 2006
1. A negative resistance R is connected to a passive network N having driving point impedance (s) as shown below. For
(s) to be positive real,
(A) |R | Re (j ) (B) |R | | (j )|
(C) |R | m (j ) (D) |R | (j ) ECE - 2007
2. For the circuit shown in the figure, the Thevenin voltage and resistance looking into X –Y are
(A) 4/3 V 2Ω (B) 4V 2/3 Ω
(C) 4/3 V 2/3Ω (D) 4V 2Ω ECE - 2009
3. In the circuit shown, what value of RL
maximizes the power delivered to RL?
(A) 2.4 Ω (B) 8 3⁄ Ω
(C) 4 Ω (D) 6 Ω
4. In the interconnection of ideal source shown in the figure, it is known that the 60V source is absorbing power.
Which of the following can be the value of the current source I?
(A) 10 A (B) 13A
(C) 15A (D) 18A
ECE - 2010
5. In the circuit shown, the power supplied by the voltage source is
(A) 0 W (B) 5 W
(C) 10 W (D) 100 W ECE - 2011
6. In the circuit shown below, the value of R such that the power transferred to R is maximum is
(A) 5Ω (B) 10Ω
(C) 15Ω (D) 20Ω
10 Ω 10 Ω
10 Ω R
5V 2V 1
1Ω
1Ω 1Ω
1Ω
1Ω
2 1
10V
1Ω
12 A + 20 V
60 V
I
4Ω 4Ω 4Ω
100V R
V
V
V
2Ω 2
1Ω i 2i 1Ω Z2(s)
Rneg
N
Z1(s)
7. In the circuit shown below, the current I is equal to
(A) 1.4 0 A (B) 2.0 0 A
(C) 2.8 0 A (D) 3.2 0 A
8. In the circuit shown below, the Norton equivalent current in amperes with respect to the terminals P and Q is
(A) 6.4 – j4.8 (B) 6.56 – j7.87
(C) 10 + j0 (D) 16 + j0
ECE/IN - 2012
9. The average power delivered to an impedance (4 j3) by a current
5cos (100t+100) A is (A) 44.2 W
(B) 50 W
(C) 62.5 W (D) 125 W ECE/EE/IN - 2012
10. The impedance looking into nodes 1 and 2 in the given circuit is
(A) 50 (B) 100
(C) 5 k
(D) 10.1 k
11. If VA VB =6 V, then VC VD is
(A) 5 V (B) 2 V
(C) 3 V (D) 6 V ECE/EE/IN - 2013
12. Consider a delta connection of resistors and its equivalent star connection as shown. If all elements of the delta connection are scaled by a factor k, k>0, the elements of the corresponding star equivalent will be scaled by a factor of
(A) k (B) k
(C) 1/k (D) √k
13. Three capacitors C1, C2 and C3 whose values are 10μF 5μF and 2μF +respectively, have breakdown voltages of 10V, 5V and 2V respectively. For the interconnection shown below , the maximum safe voltage in Volts that can be applied across the combination, and the corresponding total charge in μC stored in the effective capacitance across the terminals are respectively,
(A) 2.8 and 36 (B) 7 and 119
(C) 2.8 and 32 (D) 7 and 80 C2 C3
C1
Ra
Rb Rc
RA
RB
RC
R 10V R 2Ω
R
R R
V
R
R V
V
V 2
1Ω
5V R
2 1
99i
1kΩ
100Ω 9kΩ
i
16 0 25Ω
15Ω
j50Ω P
Q j30Ω
~
j4Ω
6Ω 6Ω 14 0 V 6Ω
j4Ω
ECE - 2014
14. For maximum power transfer between two cascaded sections of an electrical network, the relationship between the output impedance of the first section to the input impedance of the second section is
(A) (B)
(C) (D) 15. Consider the configuration shown in the
figure which is a portion of a larger electrical network
For R 1Ω and currents i =2 i 1 i 4 which one of the following is TRUE?
(A) i 5 (B) i 4
(C) Data is sufficient to conclude that the supposed currents are impossible (D) Data is insufficient to identify the
currents i i and i
16. A Y-network has resistances of 10Ω each in two of its arms, while the third arm has a resistance of 11 Ω. In the equivalent -network, the lowest value (in Ω.) among the three resistances is ________
17. Norton’s theorem states that a complex network connected to a load can be replaced with an equivalent impedance (A) in series with a current source (B) in parallel with a voltage source (C) in series with a voltage source (D) in parallel with a current source
18. In the figure shown, the value of the current I (in Amperes) is______.
19. In the circuit shown in the figure, the value of node voltage V is
(A) 22 + j 2 V (B) 2 + j 22 V
(C) 22 – j 2 V (D) 2 – j 22 V 20. The circuit shown in the figure, the
angular frequency (in rad/s) at which the Norton equivalent impedance as seen from terminals b-b′ is purely resistive is_____________.
21. For the Y-network shown in the figure, the value of R (in Ω) in the equivalent
∆-network is ____.
R
3Ω 5Ω
7.5Ω
~
b
b′
1F
0.5 1Ω
10 cos t (volts)
j6Ω V V
4 0
10 0
6Ω 4Ω j3Ω
10Ω 5Ω
5Ω
5V 1
R R R
i i
i
i i
i
22. The magnitude of current (in mA) through the resistor R in the figure shown is__________
23. The equivalent resistance in the infinite ladder network shown in the figure, is R .
The value of R /R is________
24. The circuit shown in the figure represents a
(A) voltage controlled voltage source (B) voltage controlled current source (C) current controlled current source (D) current controlled voltage source EE - 2006
1. The three limbed non ideal core shown in the figure has three winding with nominal inductances L each when measured individually with a signal phase AC source. The inductance of the winding as connected will be
(A) Very low (B) L/3
(C) 3L
(D) Very high EE - 2007
2. A 3V dc supply with an internal resistance of 2Ω supplies a passive non-linear resistance characterized by the relation VNL = . 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 EE - 2008
3. 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
4. Assuming ideal elements in the circuit shown below, the voltage Vab will be
(A) – 3 V (B) 0 V
(C) 3 V (D) 5 V EE - 2009
Statements for Linked Answer Questions 5
& 6:
2kΩ
5V
3VAB
2kΩ +
B A
+ 1kΩ
5V
b 1A
2Ω
Vab
i a
+
+ 1Ω
1Ω i
5V
4V 1Ω
1Ω 1Ω 1Ω
3Ω a b 1Ω
V +
R
R
2R R R R
R R
R R R
R 2m 4kΩ 1kΩ
2kΩ R
10 m R
R 3kΩ
5. For the circuit given above, the Thevenin’s resistance across the terminals A and B is
(A) 0.5kΩ (B) 0.2kΩ
(C) 1kΩ (D) 0.11kΩ
6. For the circuit given above, the Thevenin’s voltage across the terminals and B is
(A) 1.25V (B) 0.25V
(C) 1V (D) 0.5V
7. How many 200W/220V incandescent lamps connected in series would consume the same total power as a single 100W/220V incandescent lamp?
(A) not possible (B) 4
(C) 3 (D) 2
8. 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 Vs=20V) (B) (2A; Put Vs =8V) (C) (5A; Put Is = 10A) (D) (7A; Put Is= 12A)
9. The current through the 2 kΩ resistance in the circuit shown is
(A) 0mA (B) 1mA
(C) 2mA (D) 6mA
EE - 2010
10. If the 12 Ω 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
11. 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 EE - 2011
12. In the circuit given below, the value of R required for the transfer of maximum power to the load having a resistance of 3Ω is
(A) Zero (B) 6Ω
(C) 3Ω (D) Inifnity EC/EE/IN - 2012
13. Assuming both the voltage sources are in phase the value of R for which maximum power is transferred from circuit A to circuit B is
10 V 6Ω
3Ω oad R
Source
1Ω i
V
6V R
1Ω 2 1 12Ω
6V 1kΩ 1kΩ
1kΩ 1kΩ
2kΩ C
V 4V Is = 5A 2Ω
(A) 0.8 Ω (B) 1.4 Ω
(C) 2 Ω (D) 2.8 Ω EC/IN/EE - 2013
14. The following arrangement consists of an ideal transformer and an attenuator which attenuates by a factor of 0.8. An ac voltage V 100V is applied across WX to get an open circuit voltage V across YZ. Next , an ac voltage
V = 100V is applied across YZ to get an open circuit voltage V across WX.
Then,V /V V /V are respectively.
(A) 125/100 and 80/100 (B) 100/100 and 80/100 (C) 100/100 and 100/100 (D) 80/100 and 80/100 EE - 2014
15. The three circuit elements shown in the figure are part of an electric circuit. The total power absorbed by the three circuit elements in watts is ___________
16. In the figure, the value of resistor R is (25 + I/2) ohms, where I is the current in amperes. The current I is______
17. Two identical coupled inductors are connected in series. The measured inductances for the two possible series connections are 380 μH and 240 μH. Their mutual inductance in μH is_______
18. The total power dissipated in the circuit, shown in the figure, is 1 kW.
The voltmeter, across the load, reads 200 V. The value of is__________
19. The line A to neutral voltage is 10 15° V 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√3 105 V (B) 10 105 V
(C) 10√3 75 V (D) 10√3 90 V 20. The Norton’s equivalent source in
amperes as seen into the terminals X and Y is _______
5Ω 5Ω
5V
2.5V
2.5Ω
5Ω
~
V 200V oadR
2 1Ω 10
ac source
R 300V
8 10
2
100V 80V
15V
Y
Z W
X
1:1.25
~
j1Ω~
3V2Ω R
Circuit Circuit
10V
21. The power delivered by the current source, in the figure, is_____________
22. An incandescent lamp is marked 40 W, 240V. If resistance at room temperature (26°C) is 120 Ω, and temperature
coefficient of resistance is 4.5 10 /°C, then its ‘ON’ state filament temperature in
°C is approximately___________
IN - 2006
1. The root – mean – square value of a voltage waveform consisting of a superimposition of 2V dc and a 4V peak – to – peak square wave is
(A) 2 V (B) √6 V
(C) √8 V (D) √12 V IN - 2008
2. The power supplied by the dc voltage source in the circuit shown below is
(A) 0W (B) 1.0W
(C) 2.5W (D) 3.0W
3. Which one of the following equations is valid for the circuit shown below?
(A) 0 (B) 0 (C) 0 (D) 0
4. The current I supplied by the dc voltage source in the circuit shown below is
(A) 0A (B) 0.5A
(C) 1A (D) 2A
5. In the circuit shown below the maximum power that can be transferred to the load
is
(A) 250 W (B) 500 W
(C) 1000 W (D) 2000 W 10Ω
10 m
i(t)10√2sin
(10
00t)
1V 1A
1Ω I
1Ω
I7
5V
I4
1Ω 1Ω
1Ω 1Ω
1Ω
1Ω I5
I2
I1 I3
I6
+
1Ω 3Ω
6Ω 3Ω
3V
1Ω 1Ω 1Ω
2 1V
1V
IN - 2009
6. The source network S is connected to the load network L as shown by dashed lines.
The power transferred from S to L would be maximum when RL is
(A) 0Ω (B) 0.6 Ω
(C) 0.8 Ω (D) 2Ω IN - 2010
7. A 100 Ω , 1W resistor and a 800 Ω , 2W resistor are connected in series. The maximum dc voltage that can be applied continuously to the series circuit without exceeding the power limit of any of the resistors is
(A) 90V (B) 50 V
(C) 45 V (D) 40V IN - 2011
8. The current I shown in the circuit given below is equal to
(A) 3 A (B) 3.67 A
(C) 6 A (D) 9 A
IN - 2014
9. The circuit shown in the figure contains a dependent current source between A and terminals. The Thevenin’s equivalent resistance in kΩ between the terminals C and D is ___________.
10 V
V
5kΩ 5kΩ
10 V
C
10Ω
10Ω
10 V 10 10Ω
+
2Ω RL
3V
Load Network L Source Network S
10 V
+
Answer Keys and Explanations
ECE
1. [Ans. A]
(s) = (S) R
(s) (R R ( (s))
m( (s)) R 0 For (s)to be +ve & real,
Re ( (s)) R |Re( (s))| |R | 2. [Ans. D]
For V
Apply nodal analysis, V
2 2 V V 2i
1 0 V i V
2 2 2V 2i 0 V 2 2 V V 4V
Similarly, 2 V 4V R
V
2Ω
3. [Ans. C]
For maximum power transfer, R R
V 100V
100 8
(100 V) 4 Also V 50V
12.5 12.5 25 R V
⁄ 4Ω R 4Ω
4. [Ans. A]
In the given circuit, the current through the branch of 60 V source is (12 –I) as shown in Fig. The source of 60 V absorbs power, only if
P =(12 – I)60 is +ve. i.e., I<12.
The value of the current source, I can only be 10A given in option (a), as the currents given in other options are more than 12 A.
5. [Ans. A]
The current through all the branches are marked as shown in Fig. 1.
Apply KVL to outer loop 2( 3) 2( 2) 10 4 10 10
0
Power supplied by 10 V 10 0 0 6. [Ans. C]
For Maximum power transfer to RL R 15
10
10 10
Fig. 2 1Ω 1Ω
1Ω
1Ω
2 1
10V 1Ω
( 3)
( 2)
( 2)
3 3
Fig .1 I
60V 12 A 20 V
12 A
I 12 A
(12 – I)A +
4Ω 4Ω 4Ω
100V
V
V
x
y 2Ω 2
1Ω i
2i 1Ω
V
R R
R Thevenin’s resistance seen across the terminals of R into the rest of the network. The relevant circuit is shown in Fig.2, where the independent current source is open circuited and the voltage sources are short circuited.
R 10 (10 10) R R 15Ω
7. [Ans. B]
To find the current I in the given circuit in Fig. (1), the delta network with 6Ω each is converted to a star as shown in Fig. (2)
Then the given circuit reduces to Fig. (3), (4) and (5)
Where (2 j4)||(2 j4) 5Ω and 14 0
7 2 0
8. [Ans. A]
The Norton equivalent current is 16 0 25
(40 30) 8 tan (3
4) 8 36.86 (6.4 j4.8)
9. [Ans. B]
The load consists of a resistance and a capacitance of this, only R is passive and consumes power
So P = R
= (√ ) 4 50
*Note rms value of cos t
√ +.
10. [Ans. A]
After connecting a voltage source of V V V
(10k)( i ) 100( 99i i ) 10000i 100( 100 i )
100 10000i 20000i 100
i ( 100
20000) [ 200] V 100 99i i
100 [ 100 ( 200)]
50 R V 50
50Ω
99i V ( 99i )
V 100Ω
V 10kΩ
i
99i
1kΩ
100Ω 9kΩ
i
2 Ω 5Ω 14 0
14 0 7Ω
Fig. 4
2 Ω 2 Ω 2 Ω
j4Ω j4Ω
14 0
Fig. 3
~
14 0 V
j4 Ω
6 Ω 6 Ω
j4 Ω
Fig. 1 6 Ω
2 Ω 2 Ω
2 Ω
6 Ω 6 Ω
Fig. 2
11. [Ans. A]
For safe warping voltage across should not increased from 2V
V C V
For maximum power transfer
Load impedance = complex conjugate of
17. [Ans. D]
Norton’s theorem: Here for a complex network, after isolating the element, we short circuit the two ends and find the current . is the current of the independent current source with the equivalent resistance in parallel
18. [Ans. *] Range 0.49 to 0.51 Apply superposition theorem For voltage
5
20 0.25 mp For current
1 5
20 0.25 mp
(0.25 0.25) 0.50 mp 19. [Ans. D]
Using super node concept, we can treat nodes 1 and 2 to gather as a super node as shown in the figure by the dotted lines
Applying KCL to the super node we have 4 0 V
3j V
6 V
6j 0 . From super node we get
V V 10 0 . Solving and for V we get V 2 22j V
20. [Ans. *] Range 1.9 to 2.1
The Norton’s equivalent circuit is
In doiman
(1||0.5j ) 1 j 0.5j
1 0.5j j 0.5j (1 0.5j )
1 0.25
j
0.5j (1 0.5j ) j(1 0.25 ) (1 0.25 )
Equating the imaginary part to zero, we get
0.5 1 0.25 0 0.25 1
2rad/sec
Voltage source shorted
0.5
1F 1Ω
6jΩ 6Ω
3jΩ 4Ω 10 0 V
V2 V1
4 0
10Ω 5Ω
5Ω
1
10Ω 5Ω
5Ω
5V
R
21. [Ans. *] Range 9 to 11
R 5 3 5 7.5 3 7.5 7.5
7.5 2 5 3
7.5 10
22. [Ans. *] Range 2.79 to 2.81
By source transformation theorem
by KV 20 10k 8 0 28
10k 2.8 m 23. [Ans. *] Range 2.60 to 2.64
We know that in a infinite ladder network if all resistance are comprises of same value R then the equivalent resistance is
( √ )
. Then the above given network can be redrawn as R series with R equivalent as follows
R R 1.618R 2.618R
R
R 2.618 24. [Ans. C ]
In the above circuit in the output side there is a dependent current source which is controlled by the input current and hence it is a current control current source.
EE
1. [Ans. A]
The inductance of all three coils are ‘ ’ and they are connected to same line carrying the same current and set up the flux in the same direction
All these fluxes linked with each other and they will balance each other. So net flux will reduce drastically. Thus net L will be very low
2. [Ans. A]
3 2 1
Power delivered by source = 3×1=3W.
2Ω
3V I V
+
2 2
i i R
R (1 √5)R 2 R
R
2k 1k 4k
3k
8V 20V
R
5Ω 3Ω
7.5Ω
Power dissipated by 2Ω resistor 2 2 .
Power dissipated in non-linear element
= 3-2=1W 3. [Ans. B]
V 5 1
2 2.5V
Also, 4 V 4i 4(V V ) 4i . (1) Also, V 4V V V
V V
2 1.25 V
4(2.5 1.25) 4i (From (1)) i 1.25
4. [Ans. A]
V 2i 5 2 5 3V 5. [Ans. B]
Thevenin’s resistance is calculated using the circuit shown in fig. (1) and (2), where independent voltage source is short circuited
Write the loop equation :
V 3 V ( V 10 )10 10 V
5V 10 R V 10
5 Ω 0.2kΩ
6. [Ans. D]
To calculate thevenin’s voltage terminals A-B are kept open.
Applying source transformation into correct source
Applying source transformation current source is transformed into voltage source.
Applying KVL, {I is assumed to be in mA}
5
2 1k 3V 1k 0 5
2 2k 3V 0 (1) V
1k (2)
Put the value of in quation (1) 5
2 2V 3V 0 V 0.5V
7. [Ans. D]
For a lamp, P KV
For 200 220V⁄ lamp, K 200⁄220 Consider n lamps connected in series, Total power consumed n K 110 100 n 110 100 n 2
V
3V
1kΩ
1kΩ
5 2V
C 2kΩ 2kΩ
C
C
1kΩ V 3V
2kΩ
1kΩ 2kΩ
5V
C
2kΩ
2kΩ 1kΩ V
3 V
I
A
B Fig. 1
Fig. 1
1kΩ 1kΩ
3 V
V
A
B V 10
I
Fig. 2
( V 10)
8. [Ans. B]
The relevant circuit is shown in fig.
As the voltage across 2Ω 4V 4
2 2
In order to double the current through 2Ω resistance, V is to be doubled (Put V 8V) ]
Note that the 5 A source has no effect on the answer. However it gives 3A current through the voltage source as shown in fig.
9. [Ans. A]
As the ABCD bridge is balanced, 0 10. [Ans. B]
Current through R =1A By KVL, 1.R +6=12
R = 6 11. [Ans. B]
V i 100 and V i .1(by ohm’s law) 2i 100 i 50
12. [Ans. B]
To calculated voltage source is short circuited
R 6||R 6R 6 R
According to maximum power transfer theorem
R R 3 6R 6 R R 6Ω
13. [Ans. A]
For maximum power transfer
Source impedance = Load impedance z R
10
4 2.5 So in our circuit
pply KV in loop 5 2.5R 3 0 [R 2
2.5 0.8Ω]
14. [Ans. B]
V 100 V turns ratio V 125
V 0.8 V 100V when V 100V
100 V 125 V
X W
Y
Z V
1:1.25
~
5V~
10V
2Ω 2.5 R
3V
~
2Ω 5V2Ω 5V
10V
~
j1Ω~
3V2Ω R
Circuit Circuit 10V
R
6
R
V 4V 5 2Ω
3
Fig.
Thevenin’s circuit seen by 2 2′ will be as follow
V 100V
nd R 0.2||0.8 negligible V 100V
V V turns ratio 100V 1
1.25 80V 15. [Ans. 330]
Power absorbed by battery 100V = 100 10 1000
Power supplied by battery 80V 80 8 640
Current through 15V battery 10 8
2
Power supplied by battery 15V 15 2 30 Total power absorbed
1000 640 30 330 att
16. [Ans. 10]
300 (25 2) 600 (50 ) 10 17. [Ans. 35]
Total flux linkage = i
total
380μm
240μm 4 140μm
35μ
18. [Ans. *] Range 17.3 to 17.4 200 (10)√ R
Power dissipated (2) (1) R 10 4 10R 1000
R 996 10 400 R
400 99.6 17.33Ω 19. [Ans. C]
V lags voltage A by 90 V √3 10 90 Angle 90 with respect to A Hence
V √3 10 ( 90 15) 10√3 75
20. [Ans. 2]
Terminal x – y shorted Applying superposition i due to 5V
(5) (5 )
5 7.5 5 3
20 5 7.5
5 10 0.5 i due to 2.5V
5Ω 5V
2.5V
2.5Ω
5Ω
i 120 120
C
120
2
0.8 100V
V
2′
2.5 (2.5 2.5)
1 2 0.5
Total i 0.5 0.5 1 (NOTE: Option not matching with IIT website)
21. [Ans. 3]
R is resistance of room temperature (240)
For square wave voltage with peak – to – peak value of 4 V or amplitude = 2V, M.S.V = 4 + 4 = 8
R. . S. V √8 V
2. [Ans. D]
The given circuit in Fig. 1 is simplified as shown in Fig. 2 and Fig.3
Apply KCL at nodes,
At node, P
The circuit is shown in Fig.
Voltage across 1 =1V 1 = 1A
I+1= I1, or I+1=1 , 0 5. [Ans. B]
The circuit is shown in Fig. 1
i(t) 10√2 Sin (1000 t) 10 /
Phasor of i(t) 10√2
R j 10 j10 10 10 10 j 10
The Thevenin equivalent circuit is shown in Fig. 2
10 j 10 V⃗⃗⃗⃗ 10√2
10√2(10 j 10) 200 e
For maximum power transfer to load, 10 j 10 R 10Ω
⃗⃗⃗ V⃗⃗⃗⃗
20 10 e i (t) 10 sin( 45 )
10 (rms) 10
√2 Power transferred to load, P (rms) R 100
2 10 500 6. [Ans. C]
The circuit is shown in Fig.
7
2 R
Power supplied by the 10V source, P
Power dissipated in 2 resistance, P 2 98
(2 R )
Power transferred to the network, P(R ) P P
( )
P would be maximum, if 0 70
(2 R ) 98 2
(2 R ) 0 196
2 R 70 2 R 196 70
98 35
R 98
35 2 28
35 0.8 7. [Ans. C]
Resistor 1 : 100 Ω 1 Resistor 2 : 800 Ω, 2 W
Maximum current that resistor can withstand
√ 1 100
1 10 Similarly,
√ 2 800
1 20
If these two resistors are connected in series.
Then maximum value of
100Ω 800Ω
V I
+
2Ω RL
3V
Load Network L Source Network S
10 V I
+
+
V⃗⃗
Fig. 2 10Ω
10 m
10Ω i(t)
V (100 800) 1
20(900) 45Volts 8. [Ans. A]
The given circuit is Fig. 1
Convert the (10 V, 10 Ω) voltage source across A, B to the left into a current source, (1 A, 10 Ω). The resultant circuit is shown in Fig. 2
The circuit is further simplified as shown in Fig. 3 and 4
9 5
15 3
9. [Ans. 20]
For R calculation Independent voltage source should be short circuited So
So *R + Apply KCL at Node A 10 V 1 V
5k 10 V 5K 1 V .5 V
2V 1.5 V 0.75V So 1 V
5k
1 0.75 5k 0.25
5k 5 10 0.05m
[R 1
0.05m 20kΩ]
10 V
R 5k
5K
so V 1 C
1V( ssume) 10 V
5kΩ 5kΩ
10 V
C
V
5Ω 10Ω
Fig. 4 9
5Ω 10 10Ω
Fig. 3 1
10Ω 10 10Ω
Fig. 2 10Ω
1 10Ω
10Ω
10 V 10 10Ω
Fig. 1