Question

A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the capacitor is zero. If a 24-volt battery is connected to the circuit and the circuit is closed at t = 0, determine the charge on the capacitor at any time t.

Answer 1

Answer:

$q_c(t) = C_1e^{(-4\times 10^3)t}+C_2e^{(- 10^3)t}$

Step-by-step explanation:

given,  

C = 0.25 × 10⁻⁶ F  

R = 5 × 10³ F  

L = 1 H  

E = 24  

the equation of the circuit  

$L\dfrac{d^2q}{dt^2}+R\dfrac{dq}{dt}+\dfrac{q}{C}=E$  

substitute the given data  

$(1)\dfrac{d^2q}{dt^2}+(5 \times 10^3)\dfrac{dq}{dt}+\dfrac{q}{0.25\times 10^6}=24$  

$\dfrac{d^2q}{dt^2}+(5 \times 10^3)\dfrac{dq}{dt}+(4\times 10^6)q=24$  

auxiliary equation  

$m^2+(5 \times 10^3)+(4\times 10^6) = 0  $

$m = \dfrac{-5\times 10^3 \pm \sqrt{(5\times 10^3)^2-4\times 4\times 10^6}}{2}$

$m = \dfrac{-5\times 10^3 \pm \sqrt{25\times 10^6-16 10^6}}{2}$

$m = -4 \times 10^3,-10^3$

so, complimentary function

$q_c(t) = C_1e^{(-4\times 10^3)t}+C_2e^{(- 10^3)t}$

charge at any time t is given by the above equation.

C₁ and C₂ are arbitrary constant

which will be calculated using boundary condition.