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184_notes:examples:week8_resistors_parallel [2017/10/11 01:03] – tallpaul | 184_notes:examples:week8_resistors_parallel [2021/06/28 23:51] (current) – [Solution] schram45 | ||
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- | =====Example: | + | [[184_notes: |
+ | |||
+ | ===== Example: Resistors in Series | ||
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_series.png?300 |Circuit with Resistors in Series}} | + | [{{ 184_notes:8_res_parallel.png?300 |Circuit with Resistors in Series |
===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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* We represent the equivalent resistance of multiple resistors arranged in series as | * We represent the equivalent resistance of multiple resistors arranged in series as | ||
\begin{align*} | \begin{align*} | ||
- | R_{eq} = R_1+R_2+R_3+\ldots &&&&&& | + | R_{\text{equiv, |
\end{align*} | \end{align*} | ||
* We represent the equivalent resistance of multiple resistors arranged in parallel as | * We represent the equivalent resistance of multiple resistors arranged in parallel as | ||
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====Solution==== | ====Solution==== | ||
+ | {{184_notes: | ||
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): | ||
$$R_{\text{equiv, | $$R_{\text{equiv, | ||
+ | {{ 184_notes: | ||
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: | ||
$$R_{\text{equiv, | $$R_{\text{equiv, | ||
This yields $R_{\text{equiv, | This yields $R_{\text{equiv, | ||
- | Notice that this " | + | Notice that this " |
$$\frac{1}{R_{\text{equiv, | $$\frac{1}{R_{\text{equiv, | ||
We can plug in what we know and solve for the resistance of Resistor 2: | We can plug in what we know and solve for the resistance of Resistor 2: | ||
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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 in the circuit | + | |
+ | [{{ 184_notes: | ||
+ | A couple applications of the Loop Rule should help us find the rest of the unknowns. Consider the loop highlighted | ||
$$\Delta V_{\text{bat}}-\Delta V_1 - \Delta V_3 - \Delta V_4 = 0$$ | $$\Delta V_{\text{bat}}-\Delta V_1 - \Delta V_3 - \Delta V_4 = 0$$ | ||
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 in the circuit | + | [{{ 184_notes: |
+ | Now, consider the (different!) loop highlighted | ||
$$\Delta V_2-\Delta V_3 - \Delta V_4 = 0$$ | $$\Delta V_2-\Delta V_3 - \Delta V_4 = 0$$ | ||
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. |