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184_notes:examples:week8_wheatstone [2017/10/12 11:48] – [Solution (Part B)] tallpaul | 184_notes:examples:week8_wheatstone [2021/07/22 18:28] (current) – schram45 | ||
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===== The Wheatstone Bridge ===== | ===== The Wheatstone Bridge ===== | ||
Suppose you have the following circuit -- it is similar to a well known circuit called [[https:// | Suppose you have the following circuit -- it is similar to a well known circuit called [[https:// | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
===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 circuit is in a steady state. | + | |
- | * 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 | + | * Approximating the battery as a mechanical battery: This means the battery will supply a steady power source to the circuit |
===Representations=== | ===Representations=== | ||
* We represent [[184_notes: | * We represent [[184_notes: | ||
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\Delta V_1 = \Delta V_2, &&&&& | \Delta V_1 = \Delta V_2, &&&&& | ||
\end{align*} | \end{align*} | ||
+ | Here are the loops: | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
A simple application of Ohm's Law changes these equations into | A simple application of Ohm's Law changes these equations into | ||
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I_3 R_3 = I_4 R_4 &&&&& | I_3 R_3 = I_4 R_4 &&&&& | ||
\end{align*} | \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 $R_4$: | 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 $R_4$: | ||
$$R_4 = \frac{R_2}{R_1}R_3 = 100 \Omega$$ | $$R_4 = \frac{R_2}{R_1}R_3 = 100 \Omega$$ | ||
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The direction of current shouldn' | The direction of current shouldn' | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
- | To find $\Delta V_{\text{light}}$, | + | To find $\Delta V_{\text{light}}$, |
- | {{ 184_notes: | + | [{{ 184_notes: |
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: | ||
\begin{align*} | \begin{align*} | ||
- | I &= I_1 + I_2 & | + | I &= I_1 + I_2 & |
- | I &= I_3 + I_4 & | + | I &= I_3 + I_4 & |
\Delta V_{\text{bat}} &= \Delta V_1 + \Delta V_3 & | \Delta V_{\text{bat}} &= \Delta V_1 + \Delta V_3 & | ||
\Delta V_{\text{bat}} &= \Delta V_2 - \Delta V_{\text{light}} + \Delta V_3 & | \Delta V_{\text{bat}} &= \Delta V_2 - \Delta V_{\text{light}} + \Delta V_3 & | ||
\Delta V_{\text{light}} + \Delta V_4 &= \Delta V_3 & | \Delta V_{\text{light}} + \Delta V_4 &= \Delta V_3 & | ||
\end{align*} | \end{align*} | ||
- | If we combine | + | 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: | ||
$$\frac{\Delta V_1}{R_1} + \frac{\Delta V_2}{R_2} = \frac{\Delta V_3}{R_3} + \frac{\Delta V_4}{R_4}$$ | $$\frac{\Delta V_1}{R_1} + \frac{\Delta V_2}{R_2} = \frac{\Delta V_3}{R_3} + \frac{\Delta V_4}{R_4}$$ | ||
- | Ultimately, we wish to express $\Delta V_{\text{light}}$ in terms of our known resistances, | + | Ultimately, we wish to express $\Delta V_{\text{light}}$ in terms of our known resistances, |
- | First, we use the equations from Loops 1 and 3 to rewrite our most recent result: | + | 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: |
$$\frac{\Delta V_{\text{bat}} - \Delta V_3}{R_1} + \frac{\Delta V_2}{R_2} = \frac{\Delta V_3}{R_3} + \frac{\Delta V_3 - \Delta V_{\text{light}}}{R_4}$$ | $$\frac{\Delta V_{\text{bat}} - \Delta V_3}{R_1} + \frac{\Delta V_2}{R_2} = \frac{\Delta V_3}{R_3} + \frac{\Delta V_3 - \Delta V_{\text{light}}}{R_4}$$ | ||
- | We can also use the result | + | We can also use the equation |
$$\frac{\Delta V_{\text{bat}} - \Delta V_3}{R_1} + \frac{\Delta V_{\text{bat}} + \Delta V_{\text{light}} - \Delta V_3}{R_2} = \frac{\Delta V_3}{R_3} + \frac{\Delta V_3 - \Delta V_{\text{light}}}{R_4}$$ | $$\frac{\Delta V_{\text{bat}} - \Delta V_3}{R_1} + \frac{\Delta V_{\text{bat}} + \Delta V_{\text{light}} - \Delta V_3}{R_2} = \frac{\Delta V_3}{R_3} + \frac{\Delta V_3 - \Delta V_{\text{light}}}{R_4}$$ | ||
- | It remains to rearrange and solve for $\Delta V_{\text{light}}$: | + | Since we know $\Delta V_3$ and all the resistances, |
- | $$\Delta V_{\text{light}} = \frac{\Delta V_3\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}\right) - \Delta V_{\text{bat}}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)}{\frac{1}{R_2} + \frac{1}{R_4}}$$ | + | $$\Delta V_{\text{light}} = \frac{\Delta V_3\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}\right) - \Delta V_{\text{bat}}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)}{\frac{1}{R_2} + \frac{1}{R_4}} = 2.37 \text{ V}$$ |
+ | |||
+ | 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. | ||