Question

The capacitor in the figure shown is initially uncharged. The switch is closed at t = 0.
a) Immediately after the switch is closed, what is the current through each resistor?
b) What is the final charge on the capacitor?

Answer 1

Concepts and reason

The concepts required to solve the given problem are voltage divider rule, resistance in series and resistance in parallel combination, and Ohm’s law.

First calculate the equivalent resistance by using resistance in series and resistance in parallel combinations and then Use Ohm’s law to calculate the current across each resistor.

Apply voltage divider rule to calculate the voltage across the capacitor and then use the expression for the charge stored on the capacitor to calculate the charge stored on the capacitor.

Fundamentals

The expression for the Ohm’s law is as follows:

$V = IR$

Here, I is the current and R is the resistance of the resistor.

The equivalent resistance of the resistors connected in series is given by,

$R = {R_1} + {R_2}$

Here, ${R_1}$ and ${R_2}$ are the resistances in the circuit.

The equivalent resistance of the resistors connected in parallel is given by,

$\begin{array}{c}\\\frac{1}{R} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}}\\\\R = \frac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}\\\end{array}$

The expression for the charge stored on the capacitor is given by,

$q = CV$

Here, C is the capacitance.

(a)

Let us consider the capacitor is shorted.

The equivalent resistance of the resistors connected in parallel is given by,

$R = \frac{{{R_2}{R_3}}}{{{R_2} + {R_3}}}$

Substitute $6.00{\rm{ }}\Omega$ for ${R_2}$ and $3.00{\rm{ }}\Omega$ for ${R_3}$ in above equation as follows:

$\begin{array}{c}\\R = \frac{{\left( {6.00{\rm{ }}\Omega } \right)\left( {3.00{\rm{ }}\Omega } \right)}}{{6.00{\rm{ }}\Omega + 3.00{\rm{ }}\Omega }}\\\\ = 2.00{\rm{ }}\Omega \\\end{array}$

The equivalent resistance of the resistors connected in series is given by,

$R' = R + {R_1}$

Substitute $8.00{\rm{ }}\Omega$ for ${R_1}$ and $2.00{\rm{ }}\Omega$ for $R$ in above equation as follows:

$\begin{array}{c}\\R' = 2.00{\rm{ }}\Omega + 8.00{\rm{ }}\Omega \\\\ = 10.0{\rm{ }}\Omega \\\end{array}$

(a.1)

According to Ohm’s law, the expression for the current is given by,

${I_1} = \frac{\varepsilon }{{R'}}$

Here, $\varepsilon$ is the voltage across the battery.

Substitute 42.0 V for $\varepsilon$ and $10.0{\rm{ }}\Omega$ for $R'$ in above equation as follows:

$\begin{array}{c}\\{I_1} = \frac{{42.0{\rm{ V}}}}{{10.0{\rm{ }}\Omega }}\\\\ = 4.2{\rm{ A}}\\\end{array}$

(a.2)

Now, using the ohm’s law, the voltage drop across the resistor ${R_1}$ is as follows:

${V_1} = {I_1}{R_1}$

Substitute $8.00{\rm{ }}\Omega$ for ${R_1}$ and 4.2 A for ${I_1}$in the equation ${V_1} = {I_1}{R_1}$.

$\begin{array}{c}\\{V_1} = \left( {4.2{\rm{ A}}} \right)\left( {8.00{\rm{ }}\Omega } \right)\\\\ = 33.6{\rm{ V}}\\\end{array}$

The potential difference across resistors ${R_2}$ and ${R_3}$ is given by,

$\begin{array}{c}\\{V_2} = {V_3}\\\\ = \varepsilon - {V_1}\\\end{array}$

Substitute 42.0 V for $\varepsilon$ and 33.6 V for ${V_1}$ in above equation as follows:

$\begin{array}{c}\\{V_2} = 42.0{\rm{ V}} - 33.6{\rm{ V}}\\\\ = {\rm{8}}{\rm{.4 V}}\\\end{array}$

The expression for the current across ${R_2}$ is given by,

${I_2} = \frac{{{V_2}}}{{{R_2}}}$

Substitute ${\rm{8}}{\rm{.4 V}}$ for ${V_2}$ and $6.00{\rm{ }}\Omega$ for ${R_2}$in above equation as follows:

$\begin{array}{c}\\{I_2} = \frac{{8.4{\rm{ V}}}}{{6.00{\rm{ }}\Omega }}\\\\ = 1.4{\rm{ A}}\\\end{array}$

(a.3)

The expression for the current across ${R_3}$ is given by,

${I_3} = \frac{{{V_3}}}{{{R_3}}}$

Substitute ${\rm{8}}{\rm{.4 V}}$ for ${V_3}$ and $3.00{\rm{ }}\Omega$ for ${R_3}$in above equation as follows:

$\begin{array}{c}\\{I_3} = \frac{{8.4{\rm{ V}}}}{{3.00{\rm{ }}\Omega }}\\\\ = 2.8{\rm{ A}}\\\end{array}$

(b)

The expression for the charge stored on the capacitor is given by,

$q = C{V_C}$

Here, ${V_C}$ is the voltage across the capacitor.

According to Voltage divider rule, the voltage across the capacitor is given by,

${V_C} = \varepsilon \left( {\frac{{{R_2}}}{{{R_1} + {R_2}}}} \right)$

Substitute $42.0{\rm{ V}}$for $\varepsilon$, $8.00{\rm{ }}\Omega$ for ${R_1}$, and $6.00{\rm{ }}\Omega$ for ${R_2}$ in above equation as follows:

$\begin{array}{c}\\{V_C} = \left( {42.0{\rm{ V}}} \right)\left( {\frac{{6.00{\rm{ }}\Omega }}{{8.00{\rm{ }}\Omega + 6.00{\rm{ }}\Omega }}} \right)\\\\ = 18{\rm{ V}}\\\end{array}$

Substitute $18{\rm{ V}}$ for ${V_C}$ and $4.00{\rm{ }}\mu {\rm{F}}$ for C in above equation as follows:

$\begin{array}{c}\\q = \left( {4.00{\rm{ }}\mu {\rm{F}}} \right)\left( {18{\rm{ V}}} \right)\\\\ = 72{\rm{ }}\mu {\rm{C}}\\\end{array}$

Ans: Part a.1

The current through the resistor ${R_1}$ is $4.2{\rm{ A}}$.

Part a.2

The current through the resistor ${R_2}$ is $1.4{\rm{ A}}$.

Part a.2

The current through the resistor ${R_3}$ is $2.8{\rm{ A}}$.

Part b

The charge on the capacitor is $72{\rm{ }}\mu {\rm{C}}$.