Please answer all parts please if you can ? Thank you so much. across each element (V = |*R) and using KVL to formulate an equation. KVL states that thesum of voltages in a closed loop is always equal to zero.For first loop with current i1,-10 + 400 i, + 200( i, – iz) + 10 = 0The (-10) is the voltage from the voltage source on the left. The negative sign indicatesthe flow of current from negative to positive terminal. The (+10) indicates the voltage ofthe source on the right. The positive sign is for the flow of current from positive tonegative terminal.For second l0oop with current i2,- 10 + 200( iz - i1) + 500 iz + 100( iz – i3) = 0For the third loop with current i3, the current source dominates and the current i3 isequal to the current provided by the current source.iz = - 0.1 AThe negative sign indicates that the 100 mA current is opposite in direction to theassumed current i3.Thus, there are 2 unknowns and 2 equations to solve linearly. Find the values of i and i2.The current across the 200 ohm resistor is i1 - iz in the downward direction. If theyspecify the current flowing from bottom to top, then the subtraction is inversed (i2 – i1). Problem 1:Vị 4002 v500Ω v200210V31002 (1)100mA10VNode 0Node-Voltage method:1. Identify NodesHere, V1, V2 and V3 are the voltage nodes. We calculate currents entering and exiting aparticular node.2. Calculate current equations for each node:For node V1.V1 = 10 V as it is connected to the voltage source and hence no currents arecalculated for it.For node V2, considering V2 calculate currents for all branches connected to it using Ohm'slaw (I= V/R) and KCL.v2 - vi v2 - V3 , v2 - 10= 01.400500200If you notice the equation, the node in question (V2) is always the first node and the termis the voltage difference between two nodes wherever a component (resistor) existsbetween them. The equation is an application of KCL which states the sum of all currentsentering or exiting a particular node is always zero.For node 3,V3 - v2 V3 - 0100The (-0.1) value is the branch current for the branch with the 100 mA sourceconnected to it. If a current source is connected in a branch, it dominates and thecurrent in that branch is equal to that provided by the source. Until now we haveconsidered all the currents exiting the node to be positive (node in question isalways first term). The 100 mA or 0.1 A current is entering the node and hence hasa negative magnitude to denote the opposite direction of assumed direction.500(-0.1) = 03. Solve linear equations:You know the value of Vi, so there are two unknowns (Vzand Vs) and two equations. Solvethe equations linearly to achieve the value of the unknowns.4. Calculate the required parameter:In the manual, the required parameter is the current across the 200 ohm resistor.v2 - vi v2 - V3, V2 - 10= 0400500200In above equation, the highlighted term is the current flowing across the 200 ohm resistor(I200 = V200 / R). You have calculated the value of V and hence can easily calculate therequired current.Mesh Analysis:In mesh analysis, we use KVL to calculate the currents in closed loops to finally calculatethe required parameter.400250022002S100210Viz100mA10VLet's assume three currents flowing in the three closed loops in the circuit (is, iz, and is)with the assumed direction. I always assume this clockwise direction as it makesformulating the equations a little bit easier for me. Basically, we are writing the voltages

Answered: Image /qna-images/answer/453c36f8-5f22-4c44-9e23-fff389170784.jpg

across each element (V = I*R) and using KVL to formulate an equation. KVL states that the sum of voltages in a closed loop is always equal to zero. For first loop with current i1, -10 + 400 i, + 200(i – iz) + 10 = 0 The (-10) is the voltage from the voltage source on the left. The negative sign indicates the flow of current from negative to positive terminal. The (+10) indicates the voltage of the source on the right. The positive sign is for the flow of current from positive to negative terminal. For second loop with current i2, - 10 + 200( iz - i1) + 500 iz + 100( iz – i3) = 0 For the third loop with current i3, the current source dominates and the current i3 is equal to the current provided by the current source. iz = - 0.1 A The negative sign indicates that the 100 mA current is opposite in direction to the assumed current i3. Thus, there are 2 unknowns and 2 equations to solve linearly. Find the values of i, and i2. The current across the 200 ohm resistor is i1- iz in the downward direction. If they specify the current flowing from bottom to top, then the subtraction is inversed (i2 – i1).

across each element (V = I*R) and using KVL to formulate an equation. KVL states that the sum of voltages in a closed loop is always equal to zero. For first loop with current i1, -10 + 400 i, + 200(i – iz) + 10 = 0 The (-10) is the voltage from the voltage source on the left. The negative sign indicates the flow of current from negative to positive terminal. The (+10) indicates the voltage of the source on the right. The positive sign is for the flow of current from positive to negative terminal. For second loop with current i2, - 10 + 200( iz - i1) + 500 iz + 100( iz – i3) = 0 For the third loop with current i3, the current source dominates and the current i3 is equal to the current provided by the current source. iz = - 0.1 A The negative sign indicates that the 100 mA current is opposite in direction to the assumed current i3. Thus, there are 2 unknowns and 2 equations to solve linearly. Find the values of i, and i2. The current across the 200 ohm resistor is i1- iz in the downward direction. If they specify the current flowing from bottom to top, then the subtraction is inversed (i2 – i1).

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...

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