Question

A series circuit has a capacitor of 0.25 × 10⁻⁶ F, a resistor of 5 × 10³ Ω, and an inductor of 1H. The initial charge on the capacitor is zero. If a 12 V battery is connected to the circuit and the circuit is closed at t = 0, determine the charge on the capacitor at t = 0.001 s, at t = 0.01 s, and at any time t. Also determine the limiting charge as t → [infinity].

Answer 1

Answer:

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$

Explanation:

As we know that they are in series so the voltage across all three will be sum of all individual voltages

so it is given as

$V_r + V_L + V_c = V_{net}$

now we will have

$iR + L\frac{di}{dt} + \frac{q}{C} = 12 V$

now we have

$1\frac{d^2q}{dt^2} + (5 \times 10^3) \frac{dq}{dt} + \frac{q}{0.25 \times 10^{-6}} = 12$

So we will have

$q = 3\times 10^{-6} + c_1 e^{-4000 t} + c_2 e^{-1000 t}$

at t = 0 we have

q = 0

$0 = 3\times 10^{-6} + c_1 + c_2$

also we know that

at t = 0 i = 0

$0 = -4000 c_1 - 1000c_2$

$c_2 = -4c_1$

$c_1 = 1 \times 10^{-6}$

$c_2 = -4 \times 10^{-6}$

so we have

$q = (3 + e^{-4000 t} - 4 e^{-1000 t})\times 10^{-6}$

at t = 0.001 we have

$q = 1.55 \times 10^{-6} C$

at t = 0.01

$q = 2.99 \times 10^{-6} C$

at t = infinity

$q = 3 \times 10^{-6} C$