184_notes:examples:week8_resistors_parallel

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184_notes:examples:week8_resistors_parallel [2017/10/11 01:15] – [Example: Resistors in Series] tallpaul184_notes:examples:week8_resistors_parallel [2021/06/28 23:51] (current) – [Solution] schram45
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-=====Example: Resistors in Series=====+[[184_notes:r_parallel|Return to Resistors in Parallel Notes]] 
 + 
 +===== Example: Resistors in Series and in Parallel =====
 Suppose you have the following circuit. Resistors are labeled 1 through 4 and nodes in the circuit are labeled A, B, and C for convenience of reference. You know that the circuit contains a 12-Volt battery, $I_1 = 50 \text{ mA}$, $R_1=80 \Omega$, $R_3=300 \Omega$, $R_4=500 \Omega$, and $\Delta V_4=5 \text{ V}$. What are the potential differences $\Delta V_1$, $\Delta V_2$, $\Delta V_3$, and the resistance $R_2$? Suppose you have the following circuit. Resistors are labeled 1 through 4 and nodes in the circuit are labeled A, B, and C for convenience of reference. You know that the circuit contains a 12-Volt battery, $I_1 = 50 \text{ mA}$, $R_1=80 \Omega$, $R_3=300 \Omega$, $R_4=500 \Omega$, and $\Delta V_4=5 \text{ V}$. What are the potential differences $\Delta V_1$, $\Delta V_2$, $\Delta V_3$, and the resistance $R_2$?
  
-{{ 184_notes:8_res_parallel.png?300 |Circuit with Resistors in Series and Parallel}}+[{{ 184_notes:8_res_parallel.png?300 |Circuit with Resistors in Series and Parallel}}]
  
 ===Facts=== ===Facts===
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 ===Approximations & Assumptions=== ===Approximations & Assumptions===
-  * The wire has very very small resistance when compared to the other resistors in the circuit. +  *  The wire has very very small resistance when compared to the other resistors in the circuit: This allows there to be no energy loss across the wires and no potential difference across them either simplifying down the model. 
-  * The circuit is in a steady state. +  * The circuit is in a steady state: It takes a finite amount of time for a circuit to reach steady state and set up a charge gradient. Making this assumption means the current is not changing with time in any branch of the circuit. 
-  * Approximating the battery as a mechanical battery. +  * Approximating the battery as a mechanical battery: This means the battery will supply a steady power source to the circuit to keep it in steady state. 
-  * The resistors in the circuit are made of Ohmic materials.+  * The resistors in the circuit are made of Ohmic materials: Ohmic materials have a linear relationship between voltage and current, this allows us to use ohms law.
  
 ===Representations=== ===Representations===
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 ====Solution==== ====Solution====
 +{{184_notes:examples:week8_ex2_fig1.png?100  }}
 Let's start with resistance. The equivalent resistance for the entire circuit can be found with Ohm's Law -- equation (1): Let's start with resistance. The equivalent resistance for the entire circuit can be found with Ohm's Law -- equation (1):
 $$R_{\text{equiv, circuit}}=\frac{\Delta V_{\text{battery}}}{I_1} = 240\Omega$$ $$R_{\text{equiv, circuit}}=\frac{\Delta V_{\text{battery}}}{I_1} = 240\Omega$$
  
 +{{  184_notes:examples:week8_ex2_fig2.png?200}}
 We use $I_1$ since this is the current in the wire connected directly to the battery. We can break this down further to find the equivalent resistance in the chunk of the circuit containing Resistors 2, 3, and 4. This chunk and Resistor 1 are connected in series to form the resistance of the whole circuit, so we can use equation (2) to write: We use $I_1$ since this is the current in the wire connected directly to the battery. We can break this down further to find the equivalent resistance in the chunk of the circuit containing Resistors 2, 3, and 4. This chunk and Resistor 1 are connected in series to form the resistance of the whole circuit, so we can use equation (2) to write:
 $$R_{\text{equiv, circuit}} = R_1 + R_{\text{equiv, chunk with 2,3,4}}$$ $$R_{\text{equiv, circuit}} = R_1 + R_{\text{equiv, chunk with 2,3,4}}$$
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 A simple application of Node Rule -- equation (5) -- at Node C should tell us that $I_3 = I_4$. Now, we can reapply Ohm's Law to find the potential difference across Resistor 3: A simple application of Node Rule -- equation (5) -- at Node C should tell us that $I_3 = I_4$. Now, we can reapply Ohm's Law to find the potential difference across Resistor 3:
 $$\Delta V_3 = I_3R_3 = I_4R_3 = 3 \text{ V}$$ $$\Delta V_3 = I_3R_3 = I_4R_3 = 3 \text{ V}$$
-A couple applications of the Loop Rule should help us find the rest of the unknowns. Consider the loop highlighted in the circuit below. The Loop Rule -- equation (4) -- tells us that if we travel completely around the loop, we should encounter a total potential difference of 0. If we travel along the direction of conventional current (clockwise in our representation), voltage decreases, so $\Delta V_1$, $\Delta V_3$, $\Delta V_4<0$, whereas we have $\Delta V_{\text{bat}}>0$. These four potential differences form a loop, so they should add to 0:+ 
 +[{{ 184_notes:8_res_parallel_loop_1.png?300 |Loop 1 - Circuit with Resistors in Series and Parallel}}] 
 +A couple applications of the Loop Rule should help us find the rest of the unknowns. Consider the loop highlighted in the circuit above. The Loop Rule -- equation (4) -- tells us that if we travel completely around the loop, we should encounter a total potential difference of 0. If we travel along the direction of conventional current (clockwise in our representation), voltage decreases, so $\Delta V_1$, $\Delta V_3$, $\Delta V_4<0$, whereas we have $\Delta V_{\text{bat}}>0$. These four potential differences form a loop, so they should add to 0:
 $$\Delta V_{\text{bat}}-\Delta V_1 - \Delta V_3 - \Delta V_4 = 0$$ $$\Delta V_{\text{bat}}-\Delta V_1 - \Delta V_3 - \Delta V_4 = 0$$
 We know enough potential differences to find the voltage across Resistor 1: We know enough potential differences to find the voltage across Resistor 1:
 $$\Delta V_1 = \Delta V_{\text{bat}}-\Delta V_3 - \Delta V_4 =4 \text{ V}$$ $$\Delta V_1 = \Delta V_{\text{bat}}-\Delta V_3 - \Delta V_4 =4 \text{ V}$$
  
-Now, consider the (different!) loop highlighted in the circuit below. The Loop Rule tells us that if we travel completely around the loop, we should encounter a total potential difference of 0. We see that current in Resistor 2 runs opposite to the current in the other resistors if we follow the loop in one direction. We have to choose a direction for the application of the Loop Rule. If we go clockwise, voltage increases across Resistor 2, but drops across Resistors 3 and 4. So we write:+[{{ 184_notes:8_res_parallel_loop_2.png?300 |Loop 2 - Circuit with Resistors in Series and Parallel}}] 
 +Now, consider the (different!) loop highlighted in the circuit above. The Loop Rule tells us that if we travel completely around the loop, we should encounter a total potential difference of 0. We see that current in Resistor 2 runs opposite to the current in the other resistors if we follow the loop in one direction. We have to choose a direction for the application of the Loop Rule. If we go clockwise, voltage increases across Resistor 2, but drops across Resistors 3 and 4. So we write:
 $$\Delta V_2-\Delta V_3 - \Delta V_4 = 0$$ $$\Delta V_2-\Delta V_3 - \Delta V_4 = 0$$
 We know enough potential differences to find the voltage across Resistor 2: We know enough potential differences to find the voltage across Resistor 2:
 $$\Delta V_2 = \Delta V_3+\Delta V_4 = 8 \text{ V}$$ $$\Delta V_2 = \Delta V_3+\Delta V_4 = 8 \text{ V}$$
 +One way in which we can evaluate the solution here is to pick a few other loops in the circuit and make sure they are still valid. There are often times many more loops in a circuit than the solution goes through.
 +
 That's all! Note that there are a lot of ways to do this problem, but we chose an approach that showcases the power of knowing equivalent resistance for resistors in parallel, and the power of the Loop Rule. See if you can create a different method for finding the unknowns. That's all! Note that there are a lot of ways to do this problem, but we chose an approach that showcases the power of knowing equivalent resistance for resistors in parallel, and the power of the Loop Rule. See if you can create a different method for finding the unknowns.
  • 184_notes/examples/week8_resistors_parallel.1507684556.txt.gz
  • Last modified: 2017/10/11 01:15
  • by tallpaul