Question

Consider the RC circuit in the figure below. The switch was at position a for a long period of time and it is suddenly switched to position b at time t = 0.

For each statement select True or False.

1. The current through the resistor equals the current across the capacitor at all times.
2. In the instant after the switch is thrown the current across the capacitor is zero.
3. In the instant after the switch is thrown the voltage across the resistor is zero.
4. In the instant after the switch is thrown, the voltage across the capacitor is zero.

What is V_c at time t = 2.00 ms?
Use the following data : R = 503.5 Ω, C = 4.00 μF, V_b = 2.95 V.

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What is the total energy dissipated by the resistor R after the switch is thrown?

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Using the same data as before, consider the case when the switch is in position b for a long time, and then it is switched to position a at time t = 0. What is V_c at time t = 2.00 ms?

Answer 1

Concepts and reason

The concepts used in this problem are voltage across the capacitor as a function of time, current flowing in the capacitor as a function of time and energy stored in the capacitor. Use the concept of voltage across the capacitor as a function of time and current flowing in the capacitor as a function of time to state the statements as true or false.

Fundamentals

Voltage across the capacitor as a function of time:

The expression for voltage is:

$V = {V_{\rm{o}}}{e^{\frac{{ - t}}{{RC}}}}$ …… (1)

Here, ${V_{\rm{o}}}$ is the initial voltage, $t$ is the time, $R$ is the resistance and $C$ is the capacitance.

Current flowing through the capacitor as a function of time:

The expression for current is:

$I = {I_{\rm{o}}}{e^{\frac{{ - t}}{{RC}}}}$ …… (2)

Here, ${I_{\rm{o}}}$ is the initial current, $t$ is the time, $R$ is the resistance and $C$ is the capacitance.

Energy stored in the capacitor:

The expression for energy stored in the capacitor is:

$U = \frac{1}{2}C{V^2}$

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

(1.1)

In the given circuit, the resistor and the capacitor are connect in series, so the current through them is equal.

So, the statement is true.

(1.2)

The current through the capacitor varies with time.

$I = {I_{\rm{o}}}{e^{\frac{{ - t}}{{RC}}}}$

So, the current across the capacitor is not zero.

(1.3)

The voltage across the resistor is:

$V = IR$

The current across the resistor is not zero, so the voltage across the resistor is not zero.

(1.4)

The voltage is a function of time.

$V = {V_{\rm{o}}}{e^{\frac{{ - t}}{{RC}}}}$

So, the voltage is not zero across the capacitor.

(2)

The expression for voltage is:

$V = {V_{\rm{o}}}{e^{\frac{{ - t}}{{RC}}}}$

The voltage across the capacitor is:

${V_{\rm{c}}} = {V_{\rm{b}}}{e^{\frac{{ - t}}{{RC}}}}$

Substitute $2.95{\rm{ V}}$ for ${V_{\rm{b}}}$ , $2{\rm{ ms}}$ for $t$ , $503.5{\rm{ }}\Omega$ for $R$ and $4{\rm{ \mu F}}$ for $C$ .

$\begin{array}{c}\\{V_{\rm{c}}} = \left( {2.95{\rm{ V}}} \right){e^{\frac{{ - \left( {2{\rm{ ms}}} \right)}}{{\left( {503.5{\rm{ }}\Omega } \right)\left( {4{\rm{ \mu F}}} \right)}}}}\\\\ = \left( {2.95{\rm{ V}}} \right){e^{\frac{{ - \left( {2{\rm{ ms}} \times \frac{{{{10}^{ - 3}}{\rm{ s}}}}{{1{\rm{ ms}}}}} \right)}}{{\left( {503.5{\rm{ }}\Omega } \right)\left( {4{\rm{ \mu F}} \times \frac{{{{10}^{ - 6}}{\rm{ F}}}}{{1{\rm{ \mu F}}}}} \right)}}}}\\\\ = \left( {2.95{\rm{ V}}} \right){e^{ - 0.993}}\\\\ = 1.09{\rm{ V}}\\\end{array}$

(3)

The energy dissipated by the resistor is equal to the energy stored by the capacitor.

$U = \frac{1}{2}C{V^2}$

Substitute $2.95{\rm{ V}}$ for $V$ and $4{\rm{ \mu F}}$ for $C$ .

$\begin{array}{c}\\U = \frac{1}{2}\left( {4{\rm{ \mu F}}} \right){\left( {2.95{\rm{ V}}} \right)^2}\\\\ = \frac{1}{2}\left( {4{\rm{ \mu F}} \times \frac{{{{10}^{ - 6}}{\rm{ F}}}}{{1{\rm{ \mu F}}}}} \right){\left( {2.95{\rm{ V}}} \right)^2}\\\\ = 17.4 \times {10^{ - 6}}{\rm{ J}}\\\\ = {\bf{1}}{\bf{.74 \times 1}}{{\bf{0}}^{{\bf{ - 5}}}}{\bf{ J}}\\\end{array}$

(4)

The voltage across the capacitor is:

${V_{\rm{c}}}' = {V_{\rm{b}}} - {V_{\rm{c}}}$

Substitute $2.95{\rm{ V}}$ for ${V_{\rm{b}}}$ and $1.09{\rm{ V}}$ for ${V_{\rm{c}}}$ .

$\begin{array}{c}\\{V_{\rm{c}}}' = 2.95{\rm{ V}} - 1.09{\rm{ V}}\\\\ = {\bf{1}}{\bf{.86 V}}\\\end{array}$

Ans: Part 1.1

The statement 1 is ‘true’.