Question

a 20 resistor is connected in series with capacitor 0.01 farad and emf E volts is given by 40 e-³t +20 e-⁶t then form the differential equations​

Answer 1

Answer:

farad and emf E volts is given by 40 e-³t +20 e-⁶t then form the differential equations ...

Answer 2

Answer:

$Q=\frac{54}{55}e^\frac{-t}{2} -\frac{4}{5} e^-^3^t-\frac{2}{11}e^-^6^t$

Step-by-step explanation:

Given, $R=20$ and $Q=0.1$

The differential equation for the current is

$RI+\frac{Q}{C}=40e^-^3^t+20e^-^6^t$

⇒ $R\frac{dQ}{dt} +\frac{Q}{C}=40e^-^3^t+20e^-^6^t$

⇒ $(20D+10)Q=40e^-^3^t+20e^-^6^t$

⇒ $(2D+1)Q=4e^-^3^t+2e^-^6^t$ ⇒$Q=C_1e^\frac{-t}{2} -\frac{4}{5} e^-^3^t-\frac{2}{11} e^-^6^t$

At $t=0$ we have $Q=0.$ Therefore, $C_1=\frac{54}{55}.$

Therefore, $Q=\frac{54}{55}e^\frac{-t}{2} -\frac{4}{5} e^-^3^t-\frac{2}{11}e^-^6^t$