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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] (current) – [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 of Resistor 2: | ||
+ | $$R_2=200\Omega$$ | ||
+ | |||
+ | Okay, now for the potential differences. It will be useful in the approach we choose | ||
+ | $$I_4=\frac{\Delta V_4}{R_4}=10 \text{ mA}$$ | ||
+ | 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 | ||
+ | $$\Delta V_3 = I_3R_3 = I_4R_3 = 3 \text{ V}$$ | ||
- | We can use the Loop Rule -- equation (4) -- to find the potential difference | + | [{{ 184_notes: |
- | $$\Delta V_{bat} | + | A couple applications of the Loop Rule should help us find the rest of the unknowns. Consider the loop highlighted in the circuit above. The Loop Rule -- equation (4) -- tells us that if we travel completely around |
- | Since we know $\Delta V_{bat}$ and $\Delta V_3$, we can plug in and solve: $\Delta | + | $$\Delta V_{\text{bat}}-\Delta V_1 - \Delta |
+ | We know enough potential differences to find the voltage across Resistor 1: | ||
+ | $$\Delta V_1 = \Delta V_{\text{bat}}-\Delta V_3 - \Delta | ||
- | We can find current through | + | [{{ 184_notes: |
- | Since we know $P_1$ and $R_1$, we can plug in and solve for $I_1=\sqrt{P_1/ | + | Now, consider |
+ | $$\Delta V_2-\Delta V_3 - \Delta V_4 = 0$$ | ||
+ | We know enough potential differences to find the voltage across Resistor 2: | ||
+ | $$\Delta V_2 = \Delta V_3+\Delta V_4 = 8 \text{ | ||
+ | One way in which we can evaluate | ||
- | We now have enough information | + | That's all! Note that there are a lot of ways to do this problem, but we chose an approach that showcases |
- | $$R_{\text{1 | + | |
- | Now, equation (3) tells us $R_{eq}=R_1+R_2$, | + | |
- | The power dissipated across Resistor 2 can be found using the same rewriting of equation (2) as above: | + | |
- | $$P_2=I^2R_2= 0.6 \text{ W}$$ | + |