184_notes:examples:week8_cap_series

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184_notes:examples:week8_cap_series [2018/06/26 14:39] – curdemma184_notes:examples:week8_cap_series [2021/06/21 18:22] – schram45
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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