Differences
This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision Next revisionBoth sides next revision | ||
184_notes:examples:week8_cap_series [2017/10/11 12:56] – [Example: Resistors in Series] tallpaul | 184_notes:examples:week8_cap_series [2021/06/21 18:22] – schram45 | ||
---|---|---|---|
Line 1: | Line 1: | ||
- | =====Example: Resistors in Series===== | + | [[184_notes:c_series|Return |
- | Suppose you have the following circuit. Capacitors are labeled 1 and 2 for convenience of reference. You know that the circuit contains a 12-Volt battery, $Q_1=4.5 \mu\text{C}$, | + | |
- | {{ 184_notes: | + | =====Example: |
+ | Suppose you have the following circuit. Capacitors are labeled 1 and 2 for convenience of reference. You know that the circuit contains a 12-Volt battery, $Q_1=4.5 \mu\text{C}$, | ||
+ | |||
+ | [{{ 184_notes: | ||
===Facts=== | ===Facts=== | ||
Line 8: | Line 10: | ||
* $Q_1 = 4.5 \mu\text{C}$ | * $Q_1 = 4.5 \mu\text{C}$ | ||
* $C_2 = 0.5 \mu\text{F}$ | * $C_2 = 0.5 \mu\text{F}$ | ||
- | * $\epsilon | + | * $k = 3$ |
===Lacking=== | ===Lacking=== | ||
Line 18: | Line 20: | ||
* We are measuring things like potential differences and charges when the circuit is in a steady state. | * We are measuring things like potential differences and charges when the circuit is in a steady state. | ||
* Approximating the battery as a mechanical battery. | * Approximating the battery as a mechanical battery. | ||
+ | * Capacitors are parallel plate capacitors. | ||
===Representations=== | ===Representations=== | ||
Line 24: | Line 27: | ||
$$\frac{1}{C_{\text{equiv}}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots$$ | $$\frac{1}{C_{\text{equiv}}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots$$ | ||
* We represent the situation with diagram given above. | * We represent the situation with diagram given above. | ||
+ | |||
====Solution==== | ====Solution==== | ||
- | Shortly, we will constrain our calculations to just Resistors | + | ===Part |
+ | Let's find $C_1$. In order to use the equation for equivalent | ||
+ | $$C_{\text{equiv}}=\frac{Q_{\text{equiv}}}{\Delta V_{\text{bat}}}=0.375 \mu\text{F}$$ | ||
+ | |||
+ | Now we can solve for $C_1$ using $$\frac{1}{C_{\text{equiv}}}=\frac{1}{C_1}+\frac{1}{C_2}$$ | ||
+ | This gives us $C_1 = 1.5 \mu\text{F}$ | ||
+ | |||
+ | ===Part 2=== | ||
+ | [{{ 184_notes: | ||
- | We can use the Loop Rule -- equation (4) -- to find the potential difference across these two resistors. The potential difference across the battery has opposite sign as the differences across the resistors, if we consider the circuit as a loop of individual differences. We write: | + | Now we need to consider what happens when we insert a dielectric. It might look something like the circuit |
- | $$\Delta V_{bat} = \Delta V_1 + \Delta V_2 + \Delta V_3$$ | + | |
- | Since we know $\Delta V_{bat}$ and $\Delta V_3$, we can plug in and solve: $\Delta V_1 + \Delta V_2 = 6 \text{ V}$. | + | |
- | We can find current through | + | So when we insert |
- | Since we know $P_1$ and $R_1$, we can plug in and solve for $I_1=\sqrt{P_1/R_1}=0.1 \text{ | + | $$\frac{1}{C_{\text{equiv, |
+ | This yields | ||
+ | $$Q_{\text{new}} | ||
- | We now have enough information to find the equivalent resistance | + | **This is the charge on __both__ capacitors** since the capacitors are in series. So even if we insert a dielectric in only one of the capacitors, the charge on both will increase. |
- | $$R_{\text{1 and 2, equivalent}}=\frac{\Delta V_1 + \Delta V_2}{I}=60 \Omega$$ | + | |
- | Now, equation (3) tells us $R_{eq}=R_1+R_2$, | + | |
- | The power dissipated across Resistor 2 can be found using the same rewriting of equation (2) as above: | + | |
- | $$P_2=I^2R_2= 0.6 \text{ W}$$ | + |