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Given the circuit below where Vs=5 V, R1=2 Q, R2 = 2 Q, and R3 =2 Q, you are going to solve for 1, 12, 13, V1, V2, and V3. 11 R1 A + V1 - 13 + Vs v2 R2 V3 R3 L1 L2

Given the circuit below where Vs=5 V, R1=2 Q, R2 = 2 Q, and R3 =2 Q, you are going to solve for 1, 12, 13, V1, V2, and V3. 11 R1 A + V1 - 13 + Vs v2 R2 V3 R3 L1 L2

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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Given the circuit below where Vs=5 V, R1=2 Q, R2 = 2 Q, and R3 =2 Q, you are going to solve for 1, 12, 13, V1, V2, and V3. 11 R1 A + V1 - 12 13 V2 R2 V33 R3 Vs L1 L2 First, let's think about how many equations we'll need. We have SIX unknowns so that means we need six equations. We can use KVL, KCL and Ohm's Law to get these equations. Follow these steps to solve for the unknowns: Step 1: Write KCL at node A. Call this Eq. 1. Step 2: Write KVL around L1. Call this Eq. 2. Use 5V in this equation, not Vs. Step 3: Write KVL around L2. Call this Eq. 3. Step 4: Write Ohm's Law for V1 in terms of 1 and R1. Call this Eq. 4. Step 5: Write Ohm's Law for V2 in terms of 12 and R2. Call this Eg. 5. Step 6: Write Ohm's Law for V3 in terms of 13 and R3. Call this Eg. 6. Ok, take a deep breath. Now we are going to start plugging equations into other equations in order to ultimately get an equation that is only in terms of one variable and numeric values. Once we get a solution for our first unknown, we can use it to solve for another unknown, and then another, etc.. Step 7: Rewrite Eq. 1 in terms of voltages and resistances. This is done by plugging Eq. 4/5/6 into Eq. 1. You'll first have to rearrange Eq. 4/5/6 in terms of current (i.e., I=V/R). Make sure to keep the numbers (i.e., V1,V2,V3) clearly written so you don't mess up. You should have an equation that is in terms of V1, V2, V3, and R1, R2, R3. Go ahead and plug in the values for R1, R2, R3 so that you have more real numbers and less variable names. Let's keep calling this Eq. 1. Step 8: Now let's plug Eq. 2. into Eq. 1. using V1. You'll first have to rearrange Eq. 2 to be in the form "V1 = ..". At this point, you have eliminated V1 from Eq. 1. This is good. Your Eq. 1 should now only contain numbers and the variables V2 and V3. Step 9: Now let's plug Eq. 3. into Eq. 1 using V3. This will allow you to remove V3 from Eq. 1. and you will now have an equation that is only in terms of numbers and V2! Go ahead and solve for V2. What is the value of V2 that you found? 1.67 V (Hint: This is the right answer. I'm just going to give it to you so you get the rest right.) 2.5 V 3.33 V 5 V
Transcribed Image Text:Given the circuit below where Vs=5 V, R1=2 Q, R2 = 2 Q, and R3 =2 Q, you are going to solve for 1, 12, 13, V1, V2, and V3. 11 R1 A + V1 - 12 13 V2 R2 V33 R3 Vs L1 L2 First, let's think about how many equations we'll need. We have SIX unknowns so that means we need six equations. We can use KVL, KCL and Ohm's Law to get these equations. Follow these steps to solve for the unknowns: Step 1: Write KCL at node A. Call this Eq. 1. Step 2: Write KVL around L1. Call this Eq. 2. Use 5V in this equation, not Vs. Step 3: Write KVL around L2. Call this Eq. 3. Step 4: Write Ohm's Law for V1 in terms of 1 and R1. Call this Eq. 4. Step 5: Write Ohm's Law for V2 in terms of 12 and R2. Call this Eg. 5. Step 6: Write Ohm's Law for V3 in terms of 13 and R3. Call this Eg. 6. Ok, take a deep breath. Now we are going to start plugging equations into other equations in order to ultimately get an equation that is only in terms of one variable and numeric values. Once we get a solution for our first unknown, we can use it to solve for another unknown, and then another, etc.. Step 7: Rewrite Eq. 1 in terms of voltages and resistances. This is done by plugging Eq. 4/5/6 into Eq. 1. You'll first have to rearrange Eq. 4/5/6 in terms of current (i.e., I=V/R). Make sure to keep the numbers (i.e., V1,V2,V3) clearly written so you don't mess up. You should have an equation that is in terms of V1, V2, V3, and R1, R2, R3. Go ahead and plug in the values for R1, R2, R3 so that you have more real numbers and less variable names. Let's keep calling this Eq. 1. Step 8: Now let's plug Eq. 2. into Eq. 1. using V1. You'll first have to rearrange Eq. 2 to be in the form "V1 = ..". At this point, you have eliminated V1 from Eq. 1. This is good. Your Eq. 1 should now only contain numbers and the variables V2 and V3. Step 9: Now let's plug Eq. 3. into Eq. 1 using V3. This will allow you to remove V3 from Eq. 1. and you will now have an equation that is only in terms of numbers and V2! Go ahead and solve for V2. What is the value of V2 that you found? 1.67 V (Hint: This is the right answer. I'm just going to give it to you so you get the rest right.) 2.5 V 3.33 V 5 V
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