RI R3 R5 B 5.6 kN 1.0 k2 1.0 kN 18 V E R2 2.2 kΩ R6 1.0 k2 2.2 kΩ

Q: R5 3.ЗКО 3.3KO R3 R4 R6 R2 4.7KO 10KΩ 4.7ΚΩ R1 1KO vs+ 50V-

A: The equivalent circuit with assumed direction of current at each node is shown below: From the…

Q: 6. I= . E = 200 V R, = 150 0 R2 = 300 2 %3D

A: Given,   E = 200 VR1 = 150 ΩR2 = 300 Ω   Given circuit is,   Here,   I1 = IX + IY      ........(1)…

Q: R5 Re Ve the circuit above, VB = 2V, R; = 13kN, R6 = 41kN, then V = V art b) R2 Rs VB R33 VOut the…

A: The given operational amplifier is ideal   The current flowing into both the terminals of the…

Q: C4 CS CS C9

A: Given, Breadboard connections are

Q: Calculate the voltage of resistance R3 using superposition R2 R4 330Q 2200 330Q R3 1500 DC 10MOhm…

A:

Q: 3K R3 A R7 1K R5 6K B1 SV R1 R2 4K 4K B

A: First we need to redraw  the circuit then we can find equivalent resistance

Q: In the circuit shown below for 1=40 A.R1=460, R2=280 and K=8, find the following: R2 + vo Il RI Kix…

A: Given

Q: R5 Ω - - Ε 0 + 18 V + ΖΩ Με R3 RS6Ω Vi R₁ α + RA ΣΩΚ = [2] -6V 1

A: I am giving answer of first question as per guidelines.

Q: Isl R2 R3 ww Is2 R4 RI lo R=2kQ, R2=7kQ Ry 8kn, R=7k

A:

Q: 16 R(s) C(s), s² + 1.6s + 16 Figure 03(d)

A:

Q: (2) jlubäi For the circuit in figure below, the * :value of R1 is 2 A R1 R2 70 R3 17.50 140 V 44 Q…

A:

Q: 6. L= . E = 200 V- R, = 150 0 R2 = 300 0 %3D %3D

A:

Q: 4. R, =_2 %3D R, = 90 0 R2 = 50 R3 = 50 %3D R4 =100

A:

Q: R2 = 236 Q R3 = 244 Q Ez = E, = 3 = R = 30 Q = 1 E, = R= 160 Q E = ET = 240 V 4 = RT = IT =

A: In the circuit shown in question, the resistor R2 is in series with resistor R3 and series…

Q: In the figure shown below find the current in the resistor R1 50AN R20 Rla 16.6A O 8.3A 15V O 16.6V…

A: The current division rule is very important to find current in each branch of parallel resistance.…

Q: Q5 V1 8mV R3 1/8 K 1/5 M -12V R4 1/4 K 6mV R7 Rs 1/5 M 1/8 M V2 R1 1/8 K R2 +12 V -12 V 3 1/6 K R8…

A:

Q: 12V 30 Ω R3 5 Ω R8 + α ΑΜΕ Μ 15 Ω R₁ Μ 30 Ω R6 AMG 8 Ω R 18 V 20 Ω R4 + C 8V W 25 Ω R₂ M 50 Ω R₂ ΑΓ…

A:

Q: 4 0 R1 10 N R4 5 V R6 R2 4 N 2 S. R5 4Ω R7 R3 2S Answers : R = II

A:

Q: In the diagram below, C₁ = 6.65 MF, C₂ 1.29 MF, and C3 53.1 V. C2 C1 B C3 What is the potential…

A: capacitance- Capacitance is the ratio of the change in the electric charge of a system, to the…

Q: Q.3 Determine V out for the following figure when V= V2= 1V. R2 R6 20kO R1 R3 2k0 741 - Vout 10ka…

A:

Q: PROBLEM 5 Find Vx and Ix R2 100 4A R1 14Ω [R3 : 150 Vx 1A ·R4 20Ω: R6 :60 Figure 5 R5 Ix :50 13 6A

A:

Q: R1 S1 1K + A 0- 10mAdc R2 + 1.5K 15Vdc R4 R5 R3 3.ЗК 3.3K 1.5K R6 10K R7 10K

A:

Q: +24 V I R2 R1 4 N o +8 V R3 10 Ω |13

A:

Q: Determine the current I through R5 in Figure, when R4= 0 ohm and when R4= 50 ohm. 101 R1 20 A R Rs=…

A: Kirchhoff's voltage law: This law states that the sum of every voltage around the closed loop is…

Q: 3. CALCULATE Vout qua ป S V R3 I 2 K R₁ 5 K + R2 3K

A:

Q: A H-I X-4 240U 1552 120 VAC R1 H-2 X-3 240V. AB Figure 1 H-3 X-2 2404 102 R2 120 VBC H-4 B

A: From the circuit, the current flowing in resistor R1 is equal to line current A and the current…

Q: 6. R equivalent AO R, R2 600 600 BO Ili

A:

Q: the equivalent resistance across A R; = 60 R3 = 40 R3 = 102 R4 = 80 Rg = 60 Rea R2 = 82 R = 20 Re =…

A: RAB=10+2∥6+8∥6+4∥8+4∥8+6RAB=4+8∥6+4∥8+4∥8+6

Q: R =20 R3 =6 1 '4 V E = 6 V 1 2 R2 E3 3 V

A: Above given circuit can be solved using either of the circuit theory rules mentioned below:…

Q: 5 500mA 200mA Solve for the value of R3 as shown in the figure. 100mA 50mA 250 13R3 R1 R2 R4 12

A:

Q: 3s+25 y=C-! s2+6s+13

A:

Q: A B Vo R4 R5 6 ko 2 ko 11 R1 R2 R3 8 mA 3 ka 6 kQ 1 ka

A: Circuit is given as:

Q: R1 3kN R2 6 ΚΩ Vs O B 15y R3 8k2 FIGURE 1 From Figure 1 above find Vth (AC) O a 10 55V

A: PRESENCE OF SOURCE  In the presence of sources, all the tolerant essentials which are R, L and C…

Q: What is the total resistance RT connected to the battery, in this circuit R2 >400 120V R5 1600 R4…

A:

Q: R1 R5 R7 100 0.12 V1 100 R2 L1 10mH L3 10MH 0.10 120V 60HZ R3 R6 L2 0.10 100 10mH Figure 2.1

A:

Q: What is the equivalent resistance between points a and b when R = 22.9 Ω? R R a R R b R R Select…

A:

Q: R1 1.0Kn R4 330n R2 470 R6 1.0Kn R8 1.5Kn 100V R3 560n R7 680n 100 n R5

A: I  have attached the solution below steps

Q: 3.B) Convert the following to simplified SOP: (A’+B+C')(A'+C"+D)(B'+D')

A: Given POS expression(A'+B+C')(A'+C'+D)(B'+D')

Q: 13- 3 kN. 2ia 5 mA 1 kN O A) 3 O B) 1 O D) 5 O E) 2

A: ELECTRICAL NETWORK: When addressing a circuit problem or constructing a balanced circuit, the word…

Q: 5.6k A R3 R4 C 4.7k 6.8k R1 1k R2 1k R5 V1 5Vdc 10k V2 10 0-

A:

Q: 12Ω 6Ω V. 60 V R 24 V The value of Vo in the circuit is V. The power absorbed by the 12-Q resistor…

A: By krichoff's current law we will find out voltage then we will find out power for each resistor .

Q: Rab(C.d connected)=....2 Rab(C,d open)=.....Q a 2.52 20 b.

A: To find the equivalent resistance of a complex circuit, it is important to convert complex circuits…

Q: R1 Va R3 R5 100Q 100Q 2000 R2 2000 R4 V2 V1 12V 2000 8V 0.035 A Curren 0.0275 A t in R4 0.045 A…

A: The circuit diagram is shown below: Applying K.C.L at the node a,…

Q: R, + 2.5V + 10V 150mA R2 6V 1.5V + R3

A:

Q: Determine v(t) for t>0 in Figure Q1 (b). R3 R5 20 Ω t = 0 R1 R2 L1 C1 12V 8Ω 20 Ω 20 H 8 mF v 4 u(1)…

A:

Q: Q) Find the voltage Vs ,using superposion . R5 + V 4A R1 R2 20 R3 150 R4 E 10V E2= 5V

A:

Q: 4. L = E = 200 V= R, = 150 0 R2 = 300 N %3D 12

A:

Q: ww 5i 10 V 12 Supe Modle 22 42 32 ,v2 =-6.956 V, v3 = 0.6522 V. %3D ww

A: Here V1,V2 and V3 form super node.hence we did super node analysis and kcl.

Q: ?Calculate the current across the R1 RI=50 R2=100 R3=150 V=120v R1 R2. R3

A:

Q: +15V R4 560 Q R1 1ka R3 9100 VZR: 3V R2 3 2ka - 8V

A: The solution is given below.

Power Measurement

According to Blondel's theorem, the total power is measured with one less number of wattmeters than the number of wires present in the system. So, a single-phase and two-wire system will require one wattmeter and a single-phase and the three-wire system will require two wattmeters, and a three-phase and the three-wire system will also require two wattmeters and a three-phase and the four-wire system will require three wattmeter devices. The wattmeter is a device that measures power using one voltage and one current input as shown in the above figure. Multiple current or voltage input pair cables are used by many power analyzers and digital storage oscilloscopes (DSOs) for measuring watts, in effect within a single instrument acting as multiple wattmeters. It is possible to measure three-phase and four-wire power with one correctly specified power analyzer.