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184_notes:examples:week8_resistors_parallel [2017/10/10 20:51] – [Example: Resistors in Series] tallpaul | 184_notes:examples:week8_resistors_parallel [2021/06/28 23:51] – [Solution] schram45 | ||
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- | =====Example: Resistors in Series===== | + | [[184_notes:r_parallel|Return to Resistors in Parallel Notes]] |
- | Suppose you have the following circuit. Resistors are labeled 1 through 4 for convenience of reference. You know that the circuit contains a 12-Volt battery, and $R_1=80 \Omega$, $R_2=80 \Omega$, $I_1 = 50 \text{ mA}$, and $\Delta V_3=3 \text{ V}$. What is the resistance of and power dissipated through Resistor 4? | + | |
- | {{ 184_notes:8_res_series.png?300 |Circuit with Resistors in Series}} | + | ===== 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 $R_2$? | ||
+ | |||
+ | [{{ 184_notes:8_res_parallel.png?300 |Circuit with Resistors in Series | ||
===Facts=== | ===Facts=== | ||
- | * $\Delta V_{bat} = 12\text{ V}$ | + | * $\Delta V_{\text{bat}} = 12\text{ V}$ |
- | * $R_1=80 \Omega$ | + | |
- | * $R_2=80 \Omega$ | + | |
* $I_1 = 50 \text{ mA}$ | * $I_1 = 50 \text{ mA}$ | ||
- | * $\Delta | + | |
+ | * $R_3=300 \Omega$ | ||
+ | * $R_4=500 \Omega$ | ||
+ | | ||
===Lacking=== | ===Lacking=== | ||
- | * $R_4$, $P_4$. | + | * $\Delta V_1$, $\Delta V_2$, $\Delta V_3$, $R_2$. |
===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=== | ||
Line 24: | Line 27: | ||
\begin{align*} | \begin{align*} | ||
\Delta V = IR &&&&&& | \Delta V = IR &&&&&& | ||
- | \end{align*} | ||
- | * We represent power dissipated across a potential as | ||
- | \begin{align*} | ||
- | P = I\Delta V &&&&&& | ||
\end{align*} | \end{align*} | ||
* 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_{\text{equiv, | + | R_{\text{equiv, |
\end{align*} | \end{align*} | ||
- | * We represent the equivalent resistance of multiple resistors arranged in series | + | * We represent the equivalent resistance of multiple resistors arranged in parallel |
\begin{align*} | \begin{align*} | ||
- | \frac{1}{R_{\text{equiv, | + | \frac{1}{R_{\text{equiv, |
\end{align*} | \end{align*} | ||
* We represent the [[184_notes: | * We represent the [[184_notes: | ||
\begin{align*} | \begin{align*} | ||
- | \Delta V_1+\Delta V_2+\Delta V_3+\ldots = 0 &&&&&& | + | \Delta V_1+\Delta V_2+\Delta V_3+\ldots = 0 &&&&&& |
\end{align*} | \end{align*} | ||
- | * We represent the [[184_notes: | + | * We represent the [[184_notes: |
\begin{align*} | \begin{align*} | ||
- | \Delta V_1+\Delta V_2+\Delta V_3+\ldots = 0 &&&&&& | + | I_{\text{in}} = I_{\text{out}} |
\end{align*} | \end{align*} | ||
* We represent the situation with diagram given above. | * We represent the situation with diagram given above. | ||
+ | |||
====Solution==== | ====Solution==== | ||
- | Shortly, we will constrain our calculations | + | {{184_notes: |
+ | Let's start with resistance. The equivalent resistance for the entire circuit can be found with Ohm's Law -- equation (1): | ||
+ | $$R_{\text{equiv, circuit}}=\frac{\Delta V_{\text{battery}}}{I_1} = 240\Omega$$ | ||
+ | |||
+ | {{ 184_notes: | ||
+ | We use $I_1$ since this is the current in the wire connected directly | ||
+ | $$R_{\text{equiv, circuit}} = R_1 + R_{\text{equiv, | ||
+ | This yields $R_{\text{equiv, | ||
+ | |||
+ | Notice that this " | ||
+ | $$\frac{1}{R_{\text{equiv, | ||
+ | We can plug in what we know and solve for the resistance | ||
+ | $$R_2=200\Omega$$ | ||
- | We can use the Loop Rule -- equation (4) -- to find the potential | + | Okay, now for the potential |
- | $$\Delta | + | $$I_4=\frac{\Delta |
- | Since we know $\Delta V_{bat}$ and $\Delta V_3$, we can plug in and solve: $\Delta | + | 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: |
+ | $$\Delta | ||
- | We can find current through | + | [{{ 184_notes: |
- | Since we know $P_1$ and $R_1$, we can plug in and solve for $I_1=\sqrt{P_1/R_1}=0.1 \text{ | + | A couple applications of the Loop Rule should help us find the rest of the unknowns. Consider the loop highlighted in the circuit |
+ | $$\Delta V_{\text{bat}}-\Delta V_1 - \Delta V_3 - \Delta V_4 = 0$$ | ||
+ | We know enough potential differences to find the voltage across Resistor 1: | ||
+ | $$\Delta V_1 = \Delta V_{\text{bat}}-\Delta V_3 - \Delta V_4 =4 \text{ | ||
- | We now have enough information | + | [{{ 184_notes: |
- | $$R_{\text{1 and 2, equivalent}}=\frac{\Delta | + | Now, consider the (different!) loop highlighted in the circuit above. The Loop Rule tells us that if we travel completely around the loop, we should encounter a total potential difference of 0. We see that current in Resistor 2 runs opposite to the current in the other resistors if we follow the loop in one direction. |
- | Now, equation (3) tells us $R_{eq}=R_1+R_2$, | + | $$\Delta V_2-\Delta |
- | The power dissipated | + | We know enough potential differences to find the voltage |
- | $$P_2=I^2R_2= 0.6 \text{ | + | $$\Delta V_2 = \Delta V_3+\Delta V_4 = 8 \text{ |
+ | 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. |