Question

The switch in the figure has been in position a for a long time. It is changed to position b at t=0s.

a)What is the charge Q on the capacitor immediately after the switch is closed?

b) What the current I through the resistor immediately after the switch is closed?

c) What is the charge Q on the capacitor at t=50^-6 s ?

d)What is the current I through the resistor at t=50^-6 s?

e) What is the charge Q on the capacitor at t=200^-6 s?

f)What is the current I through the resistor at t=200^-6 s?

Answer 1

Concepts and reason

The concept required to solve this problem is Capacitance.

Initially, calculate the expression for the charge on the capacitor. Later, find the current through resistor. Finally, calculate the charge and after through the capacitor after some time.

Fundamentals

The expression for the charge enclosed by the capacitor is as follows:

$Q = CV$

Here, C is the capacitance and V is the voltage.

The current through the resistor immediately after the switch is closed using the equation ${I_0} = \frac{Q}{{RC}}$ .

Here, R is the resistance of the resistor and C is the capacitance of the capacitor.

The expression for the charge on the capacitor is as follows:

$Q = {Q_0}{e^{\frac{{ - t}}{{RC}}}}$

Here, ${Q_0}$ is the charge at time $t = 0.00{\rm{ s}}$, t is the time, R is the resistance, and C is the capacitance.

The expression for the current at time is as follows:

$I = {I_0}{e^{\frac{{ - t}}{{RC}}}}$

Here, ${I_0}$ is the current at time $t = 0.00{\rm{ s}}$.

(a)

Substitute $4.00{\rm{ }}\mu {\rm{F}}$ for C and 9.00 V for V in the equation $Q = CV$.

$\begin{array}{c}\\Q = \left( {4.00{\rm{ }}\mu {\rm{F}}} \right)\left( {9.00{\rm{ V}}} \right)\\\\ = \left( {4.00{\rm{ }}\mu {\rm{F}}} \right)\left( {\frac{{{{10}^{ - 6}}{\rm{ F}}}}{{1.00{\rm{ }}\mu {\rm{F}}}}} \right)\left( {9.00{\rm{ V}}} \right)\\\\ = 36 \times {10^{ - 6}}{\rm{ C}}\\\end{array}$

(b)

Substitute $36 \times {10^{ - 6}}{\rm{ C}}$ for Q, $25.0{\rm{ }}\Omega$ for R, and $4.00{\rm{ }}\mu {\rm{F}}$ for C in the equation ${I_0} = \frac{Q}{{RC}}$.

$\begin{array}{c}\\{I_0} = \frac{{36 \times {{10}^{ - 6}}{\rm{ C}}}}{{\left( {25.0{\rm{ }}\Omega } \right)\left( {4.00{\rm{ }}\mu {\rm{F}}} \right)}}\\\\ = \frac{{36 \times {{10}^{ - 6}}{\rm{ C}}}}{{\left( {25.0{\rm{ }}\Omega } \right)\left( {4.00{\rm{ }}\mu {\rm{F}}} \right)\left( {\frac{{{{10}^{ - 6}}{\rm{ F}}}}{{1.00{\rm{ }}\mu {\rm{F}}}}} \right)}}\\\\ = 0.36{\rm{ A}}\\\end{array}$

(c)

Substitute $36 \times {10^{ - 6}}{\rm{ C}}$ for ${Q_0}$, $50 \times {10^{ - 6}}{\rm{ s}}$for t, $25.0{\rm{ }}\Omega$ for R, and $4.00{\rm{ }}\mu {\rm{F}}$ for C in the equation $Q = {Q_0}{e^{\frac{{ - t}}{{RC}}}}$.

$\begin{array}{c}\\Q = \left( {36 \times {{10}^{ - 6}}{\rm{ C}}} \right){e^{\frac{{ - 50 \times {{10}^{ - 6}}{\rm{ s}}}}{{\left( {25.0{\rm{ }}\Omega } \right)\left( {4.00{\rm{ }}\mu {\rm{F}}} \right)}}}}\\\\ = \left( {36 \times {{10}^{ - 6}}{\rm{ C}}} \right){e^{\frac{{ - 50 \times {{10}^{ - 6}}{\rm{ s}}}}{{\left( {25.0{\rm{ }}\Omega } \right)\left( {4.00{\rm{ }}\mu {\rm{F}}} \right)\left( {\frac{{{{10}^{ - 6}}{\rm{ F}}}}{{1.00{\rm{ }}\mu {\rm{F}}}}} \right)}}}}\\\\ = 21.8 \times {10^{ - 6}}{\rm{ C}}\\\end{array}$

(d)

Substitute $50 \times {10^{ - 6}}{\rm{ s}}$for t, 0.36 A for ${I_0}$,$25.0{\rm{ }}\Omega$ for R, and $4.00{\rm{ }}\mu {\rm{F}}$ for C in the equation $I = {I_0}{e^{\frac{{ - t}}{{RC}}}}$.

$\begin{array}{c}\\I = \left( {0.36{\rm{ A}}} \right){e^{\frac{{ - 50 \times {{10}^{ - 6}}{\rm{ s}}}}{{\left( {25.0{\rm{ }}\Omega } \right)\left( {4.00{\rm{ }}\mu {\rm{F}}} \right)}}}}\\\\ = \left( {0.36{\rm{ A}}} \right){e^{\frac{{ - 50 \times {{10}^{ - 6}}{\rm{ s}}}}{{\left( {25.0{\rm{ }}\Omega } \right)\left( {4.00{\rm{ }}\mu {\rm{F}}} \right)\left( {\frac{{{{10}^{ - 6}}{\rm{ F}}}}{{1.00{\rm{ }}\mu {\rm{F }}}}} \right)}}}}\\\\ = 0.22{\rm{ A}}\\\end{array}$

(e)

Substitute $36 \times {10^{ - 6}}{\rm{ C}}$ for ${Q_0}$, $200{\rm{ }}\mu {\rm{s}}$ for t, $25.0{\rm{ }}\Omega$ for R, and $4.00{\rm{ }}\mu {\rm{F}}$ for C in the equation $Q = {Q_0}{e^{\frac{{ - t}}{{RC}}}}$.

$\begin{array}{c}\\Q = \left( {36 \times {{10}^{ - 6}}{\rm{ C}}} \right){e^{\frac{{ - 200{\rm{ \mu s}}}}{{\left( {25.0{\rm{ }}\Omega } \right)\left( {4.00{\rm{ \mu F}}} \right)}}}}\\\\ = \left( {36 \times {{10}^{ - 6}}{\rm{ C}}} \right){e^{\frac{{ - 200{\rm{ }}\mu {\rm{s}}\left( {\frac{{{{10}^{ - 6}}{\rm{ s}}}}{{1.00{\rm{ \mu s}}}}} \right)}}{{\left( {25.0{\rm{ }}\Omega } \right)\left( {4.00{\rm{ \mu F}}} \right)\left( {\frac{{{{10}^{ - 6}}{\rm{ F}}}}{{1.00{\rm{ \mu F}}}}} \right)}}}}\\\\ = 4.87 \times {10^{ - 6}}{\rm{ C}}\\\end{array}$

(f)

Substitute $200{\rm{ \mu s}}$for t, 0.36 A for ${I_0}$,$25.0{\rm{ }}\Omega$ for R, and $4.00{\rm{ }}\mu {\rm{F}}$ for C in the equation $I = {I_0}{e^{\frac{{ - t}}{{RC}}}}$.

$\begin{array}{c}\\I = \left( {0.36{\rm{ A}}} \right){e^{\frac{{ - 200{\rm{ \mu s}}}}{{\left( {25.0{\rm{ }}\Omega } \right)\left( {4.00{\rm{ }}\mu {\rm{F}}} \right)}}}}\\\\ = \left( {0.36{\rm{ A}}} \right){e^{\frac{{ - 200{\rm{ \mu s}}\left( {\frac{{{{10}^{ - 6}}{\rm{ s}}}}{{1.00{\rm{ \mu s}}}}} \right)}}{{\left( {25.0{\rm{ }}\Omega } \right)\left( {4.00{\rm{ }}\mu {\rm{F}}} \right)\left( {\frac{{{{10}^{ - 6}}{\rm{ F}}}}{{1.00{\rm{ }}\mu {\rm{F }}}}} \right)}}}}\\\\ = 0.048{\rm{ A}}\\\end{array}$

Ans: Part a

The charge on the capacitor immediately after the switch is closed is equal to$36 \times {10^{ - 6}}{\rm{ C}}$.