In the circuit below the currents are named IA la, and lc The current direction isdetermined by the source (out of positive terminal) in the midle and right branches andis clockwise in the left branch.VAB• Ia flows in the left branch of the circuit• le flows in the middle branch of the circuit• Ic flows in the right branch of the circuitAR2R3When applying KVL:R4• transverse the loop in a Clockwise direction• voltage drops are negative.R1VcVblabel the voltage across the resistorswith the same numeric subscript as the resistorR1:= 1.2k2R2= 1.8k2R3:= 1.5kNR4= 0.5k2V= 10VV= 6Va. Redraw the circuit showing current directions and polarities of the voltage acrossthe resistors.b. Write the KVL equations and apply ohms law for theloop R1-R2-R4-Vb (left Loop)loop R3-Vc-Vb-R4 (right Loop)Equation 1Equation 2c. Write the equation for KVL at node "N" Equation 3d. Put the equations in the form below and solve. if a variable does not appear in anequation, let that N =0N1A IA + N18 IB + Nc Ic = k1N24 la + N2B Ig + N2c ·Ic = k2N34 · IA+ N3B Ig + N3c Ic = k3e. Write the KVL equation for the top loop and solve for VAB

KVL - It is applied in a closed loop or mesh .The sum of all voltages in a loop is equal to the sum…

In the circuit below the currents are named Ia la, and le The current direction is determined by the source (out of positive terminal) in the middle and right branches and is clockwise in the left branch. • la flows in the left branch of the circuit • le flows in the middle branch of the circuit • l flows in the right branch of the circuit R2 R3 When applying KVL: • transverse the loop in a Clockwise direction • voltage drops are negative. R4 R1 Ve Vb label the voltage across the resistors with the same numeric subscript as the resistor R1:= 1.2k2 Rx= 1.8k2 1.5k2 R4= 0.5kN V= 10V Ve= 6V a. Redraw the circuit showing current directions and polarities of the voltage across the resistors. b. Write the KVL equations and apply ohms law for the loop R1-R2-R4-Vb (left Loop) loop R3-Vc-Vb-R4 (right Loop) Equation 1 Equation 2 c. Write the equation for KVL at node "N" Equation 3 d. Put the equations in the form below and solve. if a variable does not appear in an equation, let that N=0 N1A LA+ N18 Is + Nac Ic = k1 N24 I4 + N2B I3 + N2c lc = k2 N34 la + N35 Ig + N3c lc = k3 e. Write the KVL equation for the top loop and solve for VAB

In the circuit below the currents are named Ia la, and le The current direction is determined by the source (out of positive terminal) in the middle and right branches and is clockwise in the left branch. • la flows in the left branch of the circuit • le flows in the middle branch of the circuit • l flows in the right branch of the circuit R2 R3 When applying KVL: • transverse the loop in a Clockwise direction • voltage drops are negative. R4 R1 Ve Vb label the voltage across the resistors with the same numeric subscript as the resistor R1:= 1.2k2 Rx= 1.8k2 1.5k2 R4= 0.5kN V= 10V Ve= 6V a. Redraw the circuit showing current directions and polarities of the voltage across the resistors. b. Write the KVL equations and apply ohms law for the loop R1-R2-R4-Vb (left Loop) loop R3-Vc-Vb-R4 (right Loop) Equation 1 Equation 2 c. Write the equation for KVL at node "N" Equation 3 d. Put the equations in the form below and solve. if a variable does not appear in an equation, let that N=0 N1A LA+ N18 Is + Nac Ic = k1 N24 I4 + N2B I3 + N2c lc = k2 N34 la + N35 Ig + N3c lc = k3 e. Write the KVL equation for the top loop and solve for VAB

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

See similar textbooks

Similar questions

What will be the branch voltage when Rx is adjusted to 162? What is the magnitude and direction of the current at Rx? Applying Nodal. Rx 10 Ohm 2 A 14 Ohm 4V -12 V 3 Ohm )3 A 4 Ohm
KANA G1 500 MVA 13.8 KV x₂ = 0.2 Xo -0.1 Bara 1 SC AY 500 MVA 13.8 KV/ 500 KV x=0.12 -0 150 Bara 2 00 www 150 St G2 150 St Bara 4 Bara 3 750 MVA 500 kV/ 18 KV x=0.1 750 MVA 18 KV x"=0.18 X2 -0.18 Xo =0.09 DL 1000 MVA 500 kV/20 kV X=0.1 G3 1000 MVA 20 kV x"=0.17 X₂ -0.2 Xo =0.09 For the power system with a single line diagram; a) By choosing the base values of 1000 MVA and 20 kV in the Generator G3 region, the system is zero, Draw the reactance diagrams of the positive and negative sequence circuits. (for each line x0 = x1 = x2 = 50 ohm) b) Find the Thevenin equivalents according to the fault condition in busbar 1.
UPVOTE WILL BE GIVEN. WRITE THE COMPLETE SOLUTIONS LEGBLY. BOX THE CORRECT ANSWERS. SOLVE IN 2 DECIMAL PLACES. (*you may use KVL/KCL or any method in solving.)
1) FOR THE CIRCUIT SHOoWN, PIND THE DROP ACROSS A-B. 12 A B 22 2v - 22 1.2 1V
Question 3. Write KVL around the circuit in Figure below. Solve for Iв. RTH B IB 6.8 kN VTH 0.7 V 1.6 V
WRITE LEGIBLY AND COMPLETE WORK WITH DRAWINGS. LABEL PROPERLY. NO SHROTCUT. SKIP IF YOU ALREADY DID THIS OR ELSE DOWNVOTE
She plant eapacity of a 8-Phage Senerating 10,00 KVA station cansist of generators ot and two Yeactence 12%. each S000Kvn Senerautor ot Reactace one 18% The Semerator bus-bar fr6 m are connectud to The station which Load is taken through Three step -up Trans formmegs houning Reacience of S%%. Determine Fault MyA 500 kVA the mayimmum The ciNcut. which breaker on I. Low voltaje side and Hrsh Voltage side may have to deal with.
Determine the total current I. Determine the Voltage V on the circuit. Determine the current 12. |I, = 100 mAZ30° O 108.74272.53° mA O 6.712-33.4° V O 167.22-109.40° mA I2 O 108.742-72.53° mA O 7.614-33.4° V O 167.74-109.40° mA Z2 V 30 Ω-60 Ω 10 Ω+ 40Ω 108.474-72.53° mA O 6.174-33.4° V 162.74109.40° mA O 108.47472.53° mA O 6.712-34.3° V O 162.74-109.40° mA O O
Can you please explain the steps to get the current values i1 , i2 , i3 , i4. I already got the 4 equations needed but cannot get the correct values of the currents. Answers are on the left bottom corner.
Instructions Multiple Attempts Not allowed. This test can only be taken once. Force Completion This test can be saved and resumed later. Question Completion Status: A Moving to another question will save this response. Quèstion 2 All of the following are correct definitions except The voltage lags the current in a capacitive circuit by 90 degrees The current and voltage are in phase a purely resistive circuit The current lags the voltage in an inductive circuit by 90 degrees The current and voltage are in phase a capacitive circuit A Moving to another question will save this response.
I need help find the KVL equation on the left and right inner loop by using branch current analysis
Determine all voltages and currents in Figure 14-38 immediately after the switch is closed +VR1 + VRA iT 16 2 40 0 4 160 is 18 N "R. 10 0 VRI 60 N VR 14 0 180 V VLA