Both sides previous revision Previous revision Next revision | Previous revision Next revisionBoth sides next revision |
184_notes:examples:week8_wheatstone [2017/10/17 00:50] – dmcpadden | 184_notes:examples:week8_wheatstone [2021/06/29 00:18] – [Solution (Part B)] schram45 |
---|
| [[184_notes:combinations|Return to Larger Combinations of Resistors and Capacitors notes]] |
| |
===== The Wheatstone Bridge ===== | ===== The Wheatstone Bridge ===== |
Suppose you have the following circuit -- it is similar to a well known circuit called [[https://en.wikipedia.org/wiki/Wheatstone_bridge|a Wheatstone bridge]]. Resistors are labeled 1 through 4 for convenience of reference, and the fifth element is a light bulb, which also has some resistance. If any current at all flows through the light bulb, it will glow. You know $R_1 = 150 \Omega$, $R_2=60 \Omega$, and $R_3$ is variable, meaning you can choose its resistance. **Part A:** You find that when $R_3=250 \Omega$, the light bulb is off. What is $R_4$? **Part B:** Suppose we set $R_3=500 \Omega$, so that the light bulb glows. We also have the new information that $\Delta V_{\text{bat}} = 20 \text{ V}$, and $\Delta V_3 = 15 \text{ V}$. Which direction does conventional current flow through the light bulb? What is the voltage across the light bulb? | Suppose you have the following circuit -- it is similar to a well known circuit called [[https://en.wikipedia.org/wiki/Wheatstone_bridge|a Wheatstone bridge]]. Resistors are labeled 1 through 4 for convenience of reference, and the fifth element is a light bulb, which also has some resistance. If any current at all flows through the light bulb, it will glow. You know $R_1 = 150 \Omega$, $R_2=60 \Omega$, and $R_3$ is variable, meaning you can choose its resistance. **Part A:** You find that when $R_3=250 \Omega$, the light bulb is off. What is $R_4$? **Part B:** Suppose we set $R_3=500 \Omega$, so that the light bulb glows. We also have the new information that $\Delta V_{\text{bat}} = 20 \text{ V}$, and $\Delta V_3 = 15 \text{ V}$. Which direction does conventional current flow through the light bulb? What is the voltage across the light bulb? |
| |
{{ 184_notes:8_wheatstone.png?300 |Circuit with Wheatstone bridge}} | [{{ 184_notes:8_wheatstone.png?300 |Circuit with Wheatstone bridge}}] |
| |
FIXME Add 1-4 to the figures | |
| |
===Facts=== | ===Facts=== |
| |
===Approximations & Assumptions=== | ===Approximations & Assumptions=== |
* The wire has very very small resistance when compared to the other resistors in the circuit. | * The resistors (including the light bulb) 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. |
* The circuit is in a steady state. | * 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. |
* Approximating the battery as a mechanical battery. | * 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. |
* The resistors (including the light bulb) in the circuit are made of Ohmic materials. | * 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. |
===Representations=== | ===Representations=== |
* We represent [[184_notes:resistivity#Resistance|Ohm's Law]] as | * We represent [[184_notes:resistivity#Resistance|Ohm's Law]] as |
Here are the loops: | Here are the loops: |
| |
{{ 184_notes:8_wheatstone_small_loops.png?600 |Circuit with Wheatstone bridge and highlighted loops}} | [{{ 184_notes:8_wheatstone_small_loops.png?600 |Circuit with Wheatstone bridge and highlighted loops}}] |
| |
A simple application of Ohm's Law changes these equations into | A simple application of Ohm's Law changes these equations into |
The direction of current shouldn't be terribly hard to figure out. If current is directed to the left, it should then add via the Node Rule to the current directed down through Resistor 3. If current is directed to the right, it should then add via the Node Rule to the current directed down through Resistor 4. Before we changed $R_3$, current was zero. Now, we have increased $R_3$. We can imagine that with the increased resistance, current in Resistor 3 should decrease, so it makes sense for current to directed to the right. One could think of this result both as less current in Resistor 3, or more current in Resistor 4. There are also a lot of other arguments that would reach this same conclusion. For clearness, we have drawn the direction of conventional current in each segment below. | The direction of current shouldn't be terribly hard to figure out. If current is directed to the left, it should then add via the Node Rule to the current directed down through Resistor 3. If current is directed to the right, it should then add via the Node Rule to the current directed down through Resistor 4. Before we changed $R_3$, current was zero. Now, we have increased $R_3$. We can imagine that with the increased resistance, current in Resistor 3 should decrease, so it makes sense for current to directed to the right. One could think of this result both as less current in Resistor 3, or more current in Resistor 4. There are also a lot of other arguments that would reach this same conclusion. For clearness, we have drawn the direction of conventional current in each segment below. |
| |
{{ 184_notes:8_wheatstone_current_directions.png?300 |Circuit with Wheatstone bridge, and directed current}} | [{{ 184_notes:8_wheatstone_current_directions.png?300 |Circuit with Wheatstone bridge, and directed current}}] |
| |
To find $\Delta V_{\text{light}}$, we will need to set up some equations using the Loop Rule and Node Rule. We will focus on Nodes C and D, and the Loops highlighted below. Note that if you want to be able to solve for everything in the circuit, you have to use some loop equations and some node equations (you can't use only loop or only nodes). | To find $\Delta V_{\text{light}}$, we will need to set up some equations using the Loop Rule and Node Rule. We will focus on Nodes C and D, and the Loops highlighted below. Note that if you want to be able to solve for everything in the circuit, you have to use some loop equations and some node equations (you can't use only loop or only nodes). |
| |
{{ 184_notes:8_wheatstone_loops.png?900 |Highlighted loops in Wheatstone bridge circuit}} | [{{ 184_notes:8_wheatstone_loops.png?900 |Highlighted loops in Wheatstone bridge circuit}}] |
| |
Applying the Node Rule and the Loop Rule, we obtain the following equations: | Applying the Node Rule and the Loop Rule, we obtain the following equations: |
Since we know $\Delta V_3$ and all the resistances, all that remains is to rearrange and solve for $\Delta V_{\text{light}}$: | Since we know $\Delta V_3$ and all the resistances, all that remains is to rearrange and solve for $\Delta V_{\text{light}}$: |
$$\Delta V_{\text{light}} = \frac{\Delta V_3\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}\right) - \Delta V_{\text{bat}}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)}{\frac{1}{R_2} + \frac{1}{R_4}} = 2.37 \text{ V}$$ | $$\Delta V_{\text{light}} = \frac{\Delta V_3\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}\right) - \Delta V_{\text{bat}}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)}{\frac{1}{R_2} + \frac{1}{R_4}} = 2.37 \text{ V}$$ |
| |
| Looking at loop 2 alone between the power source and resistor 3 we would expect the voltage across any other elements in that loop to be small. Our answer agrees with this observation as 2.37 is quite small compared to both the battery and resistor 3. |
| |