Differences

This shows you the differences between two versions of the page.

--- 184_notes:examples:week8_wheatstone [2021/06/29 00:15] – schram45
+++ 184_notes:examples:week8_wheatstone [2021/07/22 18:27] – schram45
@@ Line 63: / Line 63: @@
 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}) \\
@@ Line 81: / Line 81: @@
 Since we know $\Delta V_3$ and all the resistances, all that remains is to rearrange and solve for $\Delta V_{\text{light}}$:
 $$\Delta V_{\text{light}} = \frac{\Delta V_3\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4}\right) - \Delta V_{\text{bat}}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)}{\frac{1}{R_2} + \frac{1}{R_4}} = 2.37 \text{ V}$$
+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.