Question

An emf of 100 V is applied to a series RC circuit in which the resistance is 200 ohms and the capacitance is 10^-4 farads. Determine the charge q(t) on the capacitor if q(0) = 0. Also determine the current i(t).

Answer 1

Answer on Question #70084, Physics / Other

An emf of $100\mathrm{V}$ is applied to a series RC circuit in which the resistance is 200 ohms and the capacitance is $10^{-4}$ farads. Determine the charge $q(t)$ on the capacitor if $q(0) = 0$. Also determine the current $i(t)$.

Solution:

Applying the loop law to this circuit,

$$EMF - iR - \frac{q}{C} = 0$$

where is the terminal voltage of the battery, i (lower case because it varies with time) times R is the voltage drop across the resistor, and q (lower case because it varies with time) divided by C is the potential cross the capacitor. The battery source voltage is the EMF.

Rearranging:

$$EMF = iR + \frac{q}{C}$$

We cannot solve this equation as it stands because both $i$ and $q$ are unknowns. However, they are related through calculus by the equation:

$$i = \frac{dq}{dt}$$ $$\frac{dq}{dt} + \frac{q}{RC} = \frac{emf}{R}$$

Substituting

$$\frac{dq}{dt} + \frac{10^4 q}{200} = \frac{100}{200}$$

This is a linear differential equation of first-order. The solution of is

$$q = emfC(1 - e^{(-t/RC)})$$ $$q = 100 \times 10^{-4}(1 - e^{(- \frac{t}{200 \times 10^{-4}})}$$ $$q(t) = 0.1(1 - e^{(-50t)})$$

Differentiating the above equation would yield us current.

$$i(t) = \frac{emf}{R} e^{(-t/RC)} = 0.5e^{(-50t)}$$

Answer: $q(t) = 0.1(1 - e^{(-50t)})$; $i(t) = 0.5e^{(-50t)}$

Answer provided by https://www.AssignmentExpert.com