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--- 184_notes:examples:week8_wheatstone [2017/10/11 23:39] – [Solution (Part B)] tallpaul
+++ 184_notes:examples:week8_wheatstone [2021/07/22 18:28] (current) – schram45
@@ Line 1: / Line 1: @@
+[[184_notes:combinations|Return to Larger Combinations of Resistors and Capacitors notes]]
 ===== 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 $\Delta V_{\text{bat}} = 20 \text{ V}$. 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}}]
 ===Facts===
@@ Line 9: / Line 11: @@
   * $R_3=250 \Omega$ for Part A, $R_3=500 \Omega$ for Part B.
   * $I_{\text{light}}=0$ for Part A.
-  * $\Delta V_{\text{bat}} = 20\text{ V}$ for Part B.
+  * $\Delta V_{\text{bat}} = 20\text{ V}$ and $\Delta V_3 = 15 \text{ V}$ for Part B.
 ===Lacking===
@@ Line 17: / Line 19: @@
 ===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===
   * We represent [[184_notes:resistivity#Resistance|Ohm's Law]] as
@@ Line 36: / Line 37: @@
 \Delta V_1 = \Delta V_2, &&&&& \Delta V_3 = \Delta V_4
 \end{align*}
+Here are the loops:
-{{ 184_notes:8_wheatstone_small_loops.png?300 |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
@@ Line 44: / Line 46: @@
 I_3 R_3 = I_4 R_4 &&&&& (2)
 \end{align*}
-We can refer again to there being no current in the light bulb to say more about the current in the rest of the circuit. Since there is no current in that segment, we can use the Node Rule on Nodes A and B to say $I_1=I_3$ and $I_2=I_4$, respectively. When we plug this into the equation (2), we can find that $$R_4 = \frac{I_1}{I_2}R_3$$
+We can refer again to there being no current in the light bulb to say more about the current in the rest of the circuit. Since there is no current in that segment, we can use the Node Rule on Nodes C and D to say $I_1=I_3$ and $I_2=I_4$, respectively. When we plug this into the equation (2), we can find that $$R_4 = \frac{I_1}{I_2}R_3$$
 It remains to determine the ratio between the two currents. To do this we simply rearrange equation (1) to express the ratio as being between resistors rather than currents. This gives us a final expression for $R_4$:
 $$R_4 = \frac{R_2}{R_1}R_3 = 100 \Omega$$
 ====Solution (Part B) ====
-To reiterate the new information, we just found that $R_4=100\Omega$. Now, we change Resistor 3 to $R_3=500\Omega$, and the light bulb glows, so we know current is running through it. We also know $\Delta V_{\text{bat}} = 20 \text{ V}$. We want to find which direction conventional current is directed through the light bulb, and the voltage across the light bulb.
+To reiterate the new information, we just found that $R_4=100\Omega$. Now, we change Resistor 3 to $R_3=500\Omega$, and the light bulb glows, so we know current is running through it. We also know $\Delta V_{\text{bat}} = 20 \text{ V}$ and $\Delta V_3 = 15 \text{ V}$. We want to find which direction conventional current is directed through the light bulb, and the voltage across the light bulb.
 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.
+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 A and B, 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?300 |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:
 \begin{align*}
-  I &= I_1 + I_2 &(\text{Node C}) \\
+  I &= I_1 + I_2 &(\text{Node A}) \\
-  I &= I_3 + I_4 &(\text{Node D}) \\
+  I &= I_3 + I_4 &(\text{Node B}) \\
   \Delta V_{\text{bat}} &= \Delta V_1 + \Delta V_3 &(\text{Loop 1}) \\
   \Delta V_{\text{bat}} &= \Delta V_2 - \Delta V_{\text{light}} + \Delta V_3 &(\text{Loop 2}) \\
-  \Delta V_1 + \Delta V_{\text{light}} &= \Delta V_2 &(\text{Loop 3}) \\
+  \Delta V_{\text{light}} + \Delta V_4 &= \Delta V_3 &(\text{Loop 3})
-  \Delta V_{\text{light}} + \Delta V_4 &= \Delta V_3 &(\text{Loop 4})
 \end{align*}
-If we combine the equations from Nodes C and D, and apply Ohm's Law, our resistances come into play, which we know. After applying Ohm's Law, what we have is:
+If we set the equations from Nodes C and D equal to one another then we find:
+$$I_1 + I_2=I=I_3+I_4$$
+Then, we apply Ohm's Law $I=\frac{\Delta V}{R}$, so our resistances come into play, which we know. After applying Ohm's Law, what we have is:
 $$\frac{\Delta V_1}{R_1} + \frac{\Delta V_2}{R_2} = \frac{\Delta V_3}{R_3} + \frac{\Delta V_4}{R_4}$$
-Ultimately, we wish to express $\Delta V_{\text{light}}$ in terms of our known resistances, and $\Delta V_{\text{bat}}$. We need to figure out a way to sub in for the unknowns $\Delta V_1$, $\Delta V_2$, $\Delta V_3$, and $\Delta V_4$ in terms of our known constants. Our approach, which is not the only way to solve this problem, is to express some of our unknowns in terms of $\Delta V_3$, and then try to express $\Delta V_3$ in terms of the the potential differences across the battery and light bulb.
+Ultimately, we wish to express $\Delta V_{\text{light}}$ in terms of our known resistances, and $\Delta V_{\text{bat}}$ and $\Delta V_3$. We need to figure out a way to sub in for the unknowns $\Delta V_1$, $\Delta V_2$, and $\Delta V_4$ in terms of our known constants. Our approach (which is not the only way to solve this problem) will be to solve the chosen system of equations above.
-First, we use the equations from Loops 1 and 4 to rewrite our most recent result:
-$$\frac{\Delta V_{\text{light}} - \Delta V_3}{R_1} + \frac{\Delta V_2}{R_2} = \frac{\Delta V_3}{R_3} + \frac{\Delta V_3 - \Delta V_{\text{light}}}{R_4}$$
----
-Let's start with resistance. The equivalent resistance for the entire circuit can be found with Ohm's Law -- equation (1):
-$$R_{\text{equiv, circuit}}=\frac{\Delta V_{\text{battery}}}{I_1} = 240\Omega$$
-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:
-$$R_{\text{equiv, circuit}} = R_1 + R_{\text{equiv, chunk with 2,3,4}}$$
-This yields $R_{\text{equiv, chunk with 2,3,4}}=160 \Omega$.
-Notice that this "chunk" is actually two parallel pieces of the circuit, starting at Node A and ending at Node B. The two parallel parts are Resistor 2, and then Resistors 3 and 4 together. We can use equation (3) to break down the chunk into these two pieces (we combine Resistors 3 and 4 below using equation (2)):
-$$\frac{1}{R_{\text{equiv, chunk with 2,3,4}}}= \frac{1}{R_2}+\frac{1}{R_3+R_4}$$
-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 to know the current through Resistor 4, which is found from Ohm's Law:
+First, we will use Loops 1 to solve for $\Delta V_1$ and we will use Loop 3 to solve for $\Delta V_4$. Then we can rewrite our most recent result:
-$$I_4=\frac{\Delta V_4}{R_4}=10 \text{ mA}$$
+$$\frac{\Delta V_{\text{bat}} - \Delta V_3}{R_1} + \frac{\Delta V_2}{R_2} = \frac{\Delta V_3}{R_3} + \frac{\Delta V_3 - \Delta V_{\text{light}}}{R_4}$$
-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:
+We can also use the equation from Loop 2 to sub in for $\Delta V_2$:
-$$\Delta V_3 = I_3R_3 = I_4R_3 = 3 \text{ V}$$
+$$\frac{\Delta V_{\text{bat}} - \Delta V_3}{R_1} + \frac{\Delta V_{\text{bat}} + \Delta V_{\text{light}} - \Delta V_3}{R_2} = \frac{\Delta V_3}{R_3} + \frac{\Delta V_3 - \Delta V_{\text{light}}}{R_4}$$
+Since we know $\Delta V_3$ and all the resistances, all that remains is to rearrange and solve for $\Delta V_{\text{light}}$:
+$$\Delta V_{\text{light}} = \frac{\Delta V_3\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}\right) - \Delta V_{\text{bat}}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)}{\frac{1}{R_2} + \frac{1}{R_4}} = 2.37 \text{ V}$$
-{{ 184_notes:8_res_parallel_loop_1.png?300 |Circuit with Resistors in Series and Parallel}}
+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.
-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:
-$$\Delta V_{\text{bat}}-\Delta V_1 - \Delta V_3 - \Delta V_4 = 0$$
-We know enough potential differences to find the voltage across Resistor 1:
-$$\Delta V_1 = \Delta V_{\text{bat}}-\Delta V_3 - \Delta V_4 =4 \text{ V}$$
-{{ 184_notes:8_res_parallel_loop_2.png?300 |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:
-$$\Delta V_2-\Delta V_3 - \Delta V_4 = 0$$
-We know enough potential differences to find the voltage across Resistor 2:
-$$\Delta V_2 = \Delta V_3+\Delta V_4 = 8 \text{ V}$$
-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.