Differences

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--- 184_notes:examples:week8_cap_series [2017/10/11 12:56] – [Example: Resistors in Series] tallpaul
+++ 184_notes:examples:week8_cap_series [2018/06/26 14:39] – [Solution] curdemma
@@ Line 1: / Line 1: @@
-=====Example: Resistors in Series=====
+[[184_notes:c_series|Return to capacitors in series notes]]
-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, $Q_1=4.5 \mu\text{C}$, and $C_2=0.5 \mu\text{F}$. What is the capacitance of Capacitor 1? What happens to the charge on the capacitors if we insert a dielectric material with dielectric constant $\epsilon = 3\epsilon_0$ into Capacitor 1?
-{{ 184_notes:8_cap_series.png?300 |Circuit with Capacitors in Series}}
+=====Example: Capacitors in Series=====
+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, $Q_1=4.5 \mu\text{C}$, and $C_2=0.5 \mu\text{F}$. What is the capacitance of Capacitor 1? What happens to the charge on //both// of the capacitors if we insert a dielectric material with dielectric constant $k = 3$ into Capacitor 1?
+[{{ 184_notes:8_cap_series.png?300 |Circuit with Capacitors in Series}}]
 ===Facts===
@@ Line 8: / Line 10: @@
   * $Q_1 = 4.5 \mu\text{C}$
   * $C_2 = 0.5 \mu\text{F}$
-  * $\epsilon = 3\epsilon_0$
+  * $k = 3$
 ===Lacking===
@@ Line 18: / Line 20: @@
   * We are measuring things like potential differences and charges when the circuit is in a steady state.
   * Approximating the battery as a mechanical battery.
+  * Capacitors are parallel plate capacitors.
 ===Representations===
@@ Line 24: / Line 27: @@
 $$\frac{1}{C_{\text{equiv}}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots$$
   * We represent the situation with diagram given above.
 ====Solution====
-Shortly, we will constrain our calculations to just Resistors 1 and 2. We don't have any information on Resistor 2, so our approach will be to find the equivalent resistance of 1 and 2, and then focus on just Resistor 2 using equation (3). The first steps in our approach will be to find the current and potential difference across these two resistors. //Note, this is not the only approach that would work! Another way would be to find individual potential difference across each resistor, and then focusing on Resistor 2 from there. See if you can think of yet another method...//
+===Part 1 ===
+Let's find $C_1$. In order to use the equation for equivalent capacitance of capacitors in series, as we have here, we first need the equivalent capacitance of the entire circuit. Remember that the [[184_notes:c_series#Node_Rule_and_Charge_in_Series|charge on the capacitors]] in series should be the same, so $Q_{\text{equiv}}=Q_1=Q_2$. Now, we can write:
+$$C_{\text{equiv}}=\frac{Q_{\text{equiv}}}{\Delta V_{\text{bat}}}=0.375 \mu\text{F}$$
+Now we can solve for $C_1$ using $$\frac{1}{C_{\text{equiv}}}=\frac{1}{C_1}+\frac{1}{C_2}$$
+This gives us $C_1 = 0.5 \mu\text{F}$
+===Part 2===
+[{{  184_notes:8_cap_series_dielectric.png?300|Circuit with Capacitors in Series, one capacitor has a dielectric}}]
-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 to the right. A description of what a dielectric does in a capacitor is [[184_notes:cap_charging#Dielectrics|here]]. Its [[184_notes:cap_in_cir#Finding_Capacitance|effect on capacitance]] is: $$C =\frac{k\epsilon_0 A}{d}$$
-$$\Delta V_{bat} = \Delta V_1 + \Delta V_2 + \Delta V_3$$
-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 the circuit using equations (1) and (2). We can write the power dissipated through Resistor 1 as $$P_1={I_1}^2R_1$$
+So when we insert the dielectric, we have a new capacitance for Capacitor 1: $C_{\text{1, new}}=kC_1=4.5 \mu\text{F}$. To find the new value that the capacitors are charged to, we return to the equivalent capacitance of the circuit:
-Since we know $P_1$ and $R_1$, we can plug in and solve for $I_1=\sqrt{P_1/R_1}=0.1 \text{ A}$. Recall that the [[184_notes:current#Current_in_Different_Parts_of_the_Wire|Node Rule]] tells us that the current is the same everywhere in the circuit, since the entire circuit is arranged in a series. so $I=I_1=0.1 \text{ A}$.
+$$\frac{1}{C_{\text{equiv, new}}}=\frac{1}{C_{\text{1, new}}}+\frac{1}{C_2}$$
+This yields $C_{\text{equiv, new}}=0.45 \mu\text{F}$. Now, we can find the new charge:
+$$Q_{\text{new}} = C_{\text{equiv, new}}\Delta V_{\text{bat}} = 5.4 \mu\text{C}$$
-We now have enough information to find the equivalent resistance of the two resistors, using Ohm's Law -- equation (1). We write:
+**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 $R_{eq}=R_1+R_2$, so $$R_2=R_{eq}-R_1 = 50\Omega$$
-The power dissipated across Resistor 2 can be found using the same rewriting of equation (2) as above:
-$$P_2=I^2R_2= 0.6 \text{ W}$$