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Both sides previous revision Previous revision Next revision | Previous revision | ||
184_notes:examples:week8_wheatstone [2021/06/29 00:15] – schram45 | 184_notes:examples:week8_wheatstone [2021/07/22 18:28] (current) – schram45 | ||
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[{{ 184_notes: | [{{ 184_notes: | ||
- | To find ΔVlight, | + | To find ΔVlight, |
[{{ 184_notes: | [{{ 184_notes: | ||
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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 & | ||
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Since we know ΔV3 and all the resistances, | Since we know ΔV3 and all the resistances, | ||
ΔVlight=ΔV3(1R1+1R2+1R3+1R4)−ΔVbat(1R1+1R2)1R2+1R4=2.37 V | ΔVlight=ΔV3(1R1+1R2+1R3+1R4)−ΔVbat(1R1+1R2)1R2+1R4=2.37 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. | ||