184_notes:examples:week8_cap_series

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184_notes:examples:week8_cap_series [2017/10/16 23:50] dmcpadden184_notes:examples:week8_cap_series [2021/06/21 18:22] schram45
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 +[[184_notes:c_series|Return to capacitors in series notes]]
 +
 =====Example: 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? 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}}+[{{ 184_notes:8_cap_series.png?300 |Circuit with Capacitors in Series}}]
  
 ===Facts=== ===Facts===
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 Now we can solve for $C_1$ using $$\frac{1}{C_{\text{equiv}}}=\frac{1}{C_1}+\frac{1}{C_2}$$ Now we can solve for $C_1$ using $$\frac{1}{C_{\text{equiv}}}=\frac{1}{C_1}+\frac{1}{C_2}$$
-This gives us $C_1 = 0.5 \mu\text{F}$+This gives us $C_1 = 1.5 \mu\text{F}$
  
 ===Part 2=== ===Part 2===
-{{  184_notes:8_cap_series_dielectric.png?300|Circuit with Capacitors in Series}}+[{{  184_notes:8_cap_series_dielectric.png?300|Circuit with Capacitors in Series, one capacitor has a dielectric}}]
  
 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: $$C =\frac{k\epsilon_0 A}{d}$$ 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: $$C =\frac{k\epsilon_0 A}{d}$$
  • 184_notes/examples/week8_cap_series.txt
  • Last modified: 2021/06/29 00:08
  • by schram45