Question

A capacitor with capacitance is initially charged with charge .At time , a switch is thrown to close the circuit connecting thecapacitor in series with a resistor of resistance . (Intro 1 figure)

What is the current $I_0$ that flows through the resistor immediately after theswitch is thrown?

Express your answerin terms of any or all of the quantities , ,and .

Answer 1

Concepts and reason

The main concepts used to solve the problem are capacitance and Ohm’s law.

Initially, determine the voltage across the capacitor after the switch is thrown. Finally, determine the current flowing through the resistor by using the Ohm’s law.

Fundamentals

According to Ohm’s law, the current (I) through a conductor is directly proportional to the voltage (V) across it and inversely proportional to the resistance (R) of the conductor. The expression for Ohm’s law is,

$I = \frac{V}{R}$

Here, V is the voltage, R is the resistance, and I is the current.

Charge on a capacitor is determined by using the following formula:

$Q = CV$

Here, C is the capacitance of the capacitor, Q is the charge, and V is the voltage across the capacitor.

The charge on the capacitor is,

$q = CV$

Here, C is the capacitance of the capacitor and V is the voltage across the capacitor.

Rearrange the equation for V.

$V = \frac{q}{C}$

Determine the current flowing through the resistor immediately after the switch is thrown.

According to Ohm’s law, current flowing through the resistor is given as,

${I_0} = \frac{V}{R}$

Here, V is the voltage across the resistor and R is the resistance of the resistor.

Substitute $\frac{q}{C}$ for V in equation ${I_0} = \frac{V}{R}$.

$\begin{array}{c}\\{I_0} = \frac{{\frac{q}{C}}}{R}\\\\ = \frac{q}{{RC}}\\\end{array}$

Here, q is the charge on the capacitor and C is the capacitance of the capacitor.

Ans:

The current through the resistor immediately after the switch is thrown is $\frac{q}{{RC}}$.