Chapter 8, Problem 77E

In the circuit of Fig. 8.95, a 3 mF capacitor is accidentally installed instead of the inductor. Unfortunately, that’s not the end of the problems, as it’s later determined that the real capacitor is not really well modeled by an ideal capacitor, and the dielectric has a resistance of 10 kΩ (which should be viewed as connected in parallel to the ideal capacitor). (a) Compute the circuit time constant with and without taking the dielectric resistance into account. By how much does the dielectric change your answer? (b) Calculate v_x at t = 200 ms. Does the dielectric resistance affect your answer significantly? Explain.

To determine

Find the value of circuit time constant with and without taking dielectric resistance and explain the change in time constant by dielectric resistance.

Answer to Problem 77E

Value of time constant without dielectric is $40.8\text{ms}$ value of time constant with dielectric is $40.7446\text{ms}$ and value of percentage change in time constant is $0.136\%$.

Explanation of Solution

Given Data:

Value of capacitor by which the $3\text{ mH}$ inductor is replaced is $3\text{mF}$ and

Value of dielectric resistance is $10\text{kΩ}$.

Formula used:

The expression for resistance is as follows:

$R=\frac{v}{i}$ (1)

Here,

$v$ is the voltage ,

$i$ is the current and

$R$ is value of resistance.

The expression for the circuit time constant is as follows:

$\tau =R_{eq}C$ (2)

Here,

$\tau$ is the circuit time constant,

$R_{eq}$ is the equivalent resistance across the capacitor and

$C$ is the value of capacitance.

Calculation:

To find equivalent resistance of a circuit seen by the capacitor, open circuit the inductor and replace all independent sources by their internal resistances. That is, replace current source by open circuit.

Circuit contains dependent voltage source. So to find equivalent resistance connect an arbitrary voltage source of $1\text{V}$ across open circuit terminal.

The circuit diagram is redrawn as shown in Figure 1.

Refer to the redrawn Figure 1.

Circuit is drawn for without dielectric resistor.

The expression for Kirchhoff’s voltage law in mesh $DABCD$ is as follows:

$-v_1-i_1{R}_1+v_2+v_x=0$ (3)

Here,

$v_1$ and $v_2$ are the voltage in loop $DABCD$,

$v_x$ is voltage across resistor $R_2$,

$i_1$ is the current flowing in loop $DABCD$ and

$R_1$ is the resistances in the circuit.

Substitute $0.1v_x$ for $v_1$, $10\text{Ω}$ for $R_1$ and $1\text{V}$ for $v_2$ in equation (3):

$-0.1v_x-i_1\left(10\text{Ω}\right)+1\text{V}+v_x=0$

$0.9v_x-10i_1+1\text{V}=0$ (4)

The expression for $v_x$ is as follows:

$v_x=-i_1{R}_2$ (5)

Here,

$v_x$ is voltage across resistor $R_2$,

$i_1$ is the current flowing in loop $DABCD$ and

$R_2$ is the resistances in the circuit.

Substitute $4\text{Ω}$ for $R_2$ in equation (5):

$v_x=-i_1\left(4\text{Ω}\right)$

Substitute $-i_1\left(4\text{Ω}\right)$ for $v_x$ in equation (4):

$\begin{array}{c}0.9\left(-i_1\left(4\text{Ω}\right)\right)-10i_1+1\text{V}=0\\ -3.6i_1-10i_1+1\text{V}=0\end{array}$

$-13.6i_1+1\text{V}=\text{0}$ (6)

Rearrange equation (6) for $i_1$:

$-13.6i_1=-1\text{V}$

Simplify for $i_1$:

$i_1=0.07353\text{A}$

Substitute $1\text{V}$ for $v$ and $0.07353\text{A}$ for $i$ in equation (1):

$\begin{array}{c}R=\frac{1\text{V}}{0.07353\text{A}}\\ =13.6\text{Ω}\end{array}$

Substitute $13.6\text{Ω}$ for $R_{eq}$ and $\text{3 mF}$ for $C$ in equation (2):

$\begin{array}{c}\tau =\left(13.6\text{Ω}\right)\left(\text{3 mF}\right)\\ =40.8\text{ms}\end{array}$

So, value of time constant without dielectric is $40.8\text{ms}$.

To find equivalent resistance of a circuit, independent sources are replaced by their internal resistances. Therefore, replace current source by open circuit.

Circuit is containing dependent voltage source. So, to find equivalent resistance connect a voltage source of $1\text{V}$ across open circuit terminal.

The circuit diagram is redrawn as shown in Figure 2.

Refer to the redrawn Figure 2:

Circuit is drawn for resistor with dielectric.

The expression for KVL in mesh $DABCD$ is as follows:

$-v_1-i_1{R}_1+\left(i_2-i_1\right)R_3+v_x=0$ (7)

Here,

$v_1$ is the voltage in loop $DABCD$,

$v_x$ is voltage across resistor $R_2$,

$i_1$ is the current flowing in loop $DABCD$,

$i_2$ is the current flowing in loop $FEBCF$ and

$R_1$ is the resistances in the circuit.

The expression for KVL in mesh $FEBCF$ is as follows:

$-v_2+\left(i_2-i_1\right)R_3=0$ (8)

Here,

$v_2$ is the voltage in loop $FEBCF$,

$i_1$ is the current flowing in loop $DABCD$,

$i_2$ is the current flowing in loop $FEBCF$ and

$R_3$ is the resistances in the circuit.

Substitute $0.1v_x$ for $v_1$, $10\text{kΩ}$ for $R_3$ and $10\text{Ω}$ for $R_1$ in equation (7):

$\begin{array}{l}-0.1v_x-i_1\left(10\text{Ω}\right)+\left(i_2-i_1\right)10\text{kΩ}+v_x=0\\ 0.9v_x-10i_1+\left(i_2-i_1\right)10000\text{Ω}=0\left\{\because 1\text{kΩ}=1000\text{Ω}\right\}\end{array}$

$0.9v_x-10010i_1+i_{2}10000\text{Ω}=0$ (9)

Substitute $-i_1\left(4\text{Ω}\right)$ for $v_x$ in equation (9):

$\begin{array}{l}0.9\left(-i_1\left(4\text{Ω}\right)\right)-10010i_1+i_{2}10000\text{Ω}=0\\ -3.6i_1-10010i_1+i_{2}10000\text{Ω}=0\end{array}$

$-10013.6i_1+i_{2}10000\text{Ω}=0$ (10)

Rearrange (10) for $i_2$:

$i_2=1.00136i_1$ (11)

Substitute $1\text{V}$ for $v_2$ and $10\text{kΩ}$ for $R_3$ in equation (8):

$\begin{array}{l}-1\text{V}+\left(i_2-i_1\right)10\text{kΩ}=0\\ -1\text{V}+\left(i_2-i_1\right)10000\text{Ω}=0\left\{\because 1\text{kΩ}=1000\text{Ω}\right\}\end{array}$

$-1{\text{V+10000i}}_2-10000i_1=0$ (12)

Rearrange equation (12) for $i_1$:

$i_1=\frac{-1V+10000i_2}{10000}$

Substitute $\frac{-1V+10000i_2}{10000}$ for $i_1$ in equation (11):

$i_2=1.00136\left(\frac{-1V+10000i_2}{10000}\right)$

$i_2=\frac{-1.00136+10013.6i_2}{10000}$ (13)

Rearrange equation (13) for $i_2$:

$13.6i_2=1.00136$

Simplify for $i_2$:

$i_2=0.07363\text{A}$

Substitute $1\text{V}$ for $v$ and $0.07363\text{A}$ for $i$ in equation (1):

$\begin{array}{c}R=\frac{1\text{V}}{0.07363\text{A}}\\ =13.5815\text{Ω}\end{array}$

Substitute $13.5815\text{Ω}$ for $R_{eq}$ and $\text{3 mF}$ for $C$ in equation (2):

$\begin{array}{c}\tau =\left(13.5815\text{Ω}\right)\left(\text{3 mF}\right)\\ =40.7446\text{ms}\end{array}$

So value of time constant with dielectric is $40.7446\text{ms}$.

The expression for percentage change in time constant with dielectric resistor is as follows;

$\Delta \tau =\frac{{\tau }_1-{\tau }_2}{{\tau }_1}\times 100$ (14)

Here,

$\Delta \tau$ is the percentage change in time constant,

${\tau }_1$ is the value of time constant without dielectric and

${\tau }_2$ is the value of time constant with dielectric.

Substitute $40.8\text{ms}$ for ${\tau }_1$ and $40.7446\text{ms}$ for ${\tau }_2$ in equation (14):

$\begin{array}{c}\Delta \tau =\frac{40.8\text{ms}-40.7446\text{ms}}{40.8\text{s}}\times 100\\ =\frac{0.0554\text{ms}}{40.8\text{ms}}\times 100\\ =0.136\%\end{array}$

Conclusion:

Thus, value of time constant without dielectric is $40.8\text{ms}$ value of time constant with dielectric is $40.7446\text{ms}$ and value of percentage change in time constant is $0.136\%$.

To determine

Calculate $v_x$ at $t=200\text{ms}$ and calculate change in voltage $v_x$ by dielectric resistance.

Answer to Problem 77E

Value of voltage $v_x$ without dielectric is $0.04372\text{mV}$, value of voltage $v_x$ with dielectric is $0.04938\text{mV}$ and value of percentage change in voltage $v_x$ is $11.46\%$.

Explanation of Solution

Given Data:

Value of time $t$ is $200\text{ms}$.

Formula used:

The expression for voltage is as follows:

$v=iR$ (15)

Here,

$v$ is the voltage,

$i$ is the current and

$R$ is value of resistance.

The expression for the final response is as follows:

$v_C(t)=v_C(\infty )+\left(v_C(0^{+})-v_C(\infty )\right)e^{\frac{-t}{\tau }}$ (16)

Here,

$v_C(t)$ is the voltage across the capacitor,

$v_C(\infty )$ is the voltage across the capacitor at $t=\infty$,

$v_C(0^{+})$ is the voltage across the capacitor at $t=0^{+}$ and

$t$ is the given time

The expression for current flowing through capacitor is as follows:

$i_C=C\frac{d{v}_C}{dt}$ (17)

Here,

$i_C$ is the capacitor current,

$C$ is the capacitor and

$v_C$ is the voltage across capacitor.

Calculation:

The circuit diagram is redrawn as shown in Figure 3 for $t<0$.

Refer to the redrawn Figure 3:

Circuit is drawn for resistor without dielectric.

As there is no independent source present in the circuit which means voltage across capacitor is also $0\text{V}$.

So, value of $v_C\left(0^{-}\right)$ is $0\text{V}$.

At $t=0^{+}$ the capacitor does not allow sudden change in the voltage.

Therefore,

$v_C(0^{+})=v_C(0^{-})$

So,

$v_C\left(0^{+}\right)=0\text{ V}$

The circuit diagram is redrawn as shown in Figure 4 for $t>0$.

At $t=\infty$ the capacitor in the $RC$ circuit is connected for long time.

So, capacitor behaves as open circuit.

The circuit diagram is redrawn as shown in Figure 5 for $t=\infty$.

Refer to the redrawn Figure 5:

The expression for voltage $v_1$ is as follows:

$v_1=0.1v_x$ (18)

Here,

$v_1$ is voltage in the circuit and

$v_x$ is voltage across resistor $R_2$.

As capacitor is open circuited, current flowing through resistor $R_2$ is $0\text{A}$.

Which means value of voltage $v_x$ is $0\text{V}$.

Substitute $0\text{V}$ for $v_x$ in equation (18):

$\begin{array}{c}{v}_1=0.1\left(0\text{V}\right)\\ =0\text{V}\end{array}$

The circuit diagram is redrawn as shown in Figure 6 for $t=\infty$.

Refer to the redrawn Figure 6:

Substitute $2\text{mA}$ for $i$ and $10\text{Ω}$ for $R_1$ in equation (15):

$\begin{array}{c}v=\left(2\text{mA}\right)\left(10\text{Ω}\right)\\ =\left(2\times {10}^{-3}\text{A}\right)\left(10\text{Ω}\right)\left\{\because 1\text{mA}={10}^{-3}\text{A}\right\}\\ =0.02\text{V}\end{array}$

So,

$v_C\left(\infty \right)=0.02\text{V}$

Substitute $0.02\text{V}$ for $v_C\left(\infty \right)$, $0\text{V}$ for $v_C(0^{+})$, $40.8\text{ms}$ for $\tau$ in equation (16):

$\begin{array}{c}{v}_C(t)=0.02\text{V}+\left(0\text{V}-0.02\text{V}\right)e^{\frac{-t}{40.8\text{ms}}}\\ =0.02\left(1-e^{\frac{-t}{40.8\text{ms}}}\right)\text{V}\end{array}$

Substitute $0.02\left(1-e^{\frac{-t}{40.8\text{ms}}}\right)\text{V}$ for $v_C$ and $\text{3 mF}$ for $C$ in equation (17):

$\begin{array}{c}{i}_C=\text{3 mF}\frac{d}{dt}\left(0.02\left(1-e^{\frac{-t}{40.8\text{ms}}}\right)\text{V}\right)\\ =\text{3 mF}\left(0.02\right)\left(\frac{1}{40.8\text{ms}}{e}^{\frac{-t}{40.8\text{ms}}}\text{V}\right)\end{array}$

$i_C=1.4706e^{\frac{-t}{40.8\text{ms}}}\text{mA}$ (19)

Substitute $200\text{ms}$ for $t$ in equation (19):

$\begin{array}{c}{i}_C=1.4706e^{\frac{-200\text{ms}}{40.8\text{ms}}}\text{mA}\\ =0.01093\text{mA}\end{array}$

The expression for voltage $v_x$ is as follows:

$v_x=i_C{R}_2$ (20)

Here,

$v_x$ is voltage across resistor $R_2$,

$i_C$ is capacitor current and

$R_2$ is the resistance in the circuit.

Substitute $0.01093\text{mA}$ for $i_C$ and $4\text{Ω}$ for $R_2$ in equation (20):

$\begin{array}{c}{v}_x=\left(0.01093\text{mA}\right)4\text{Ω}\\ =0.04372\text{mV}\end{array}$

So, value of voltage $v_x$ without dielectric is $0.04372\text{mV}$.

Circuit is drawn for resistor with dielectric.

The circuit diagram is redrawn as shown in Figure 7 for $t<0$.

Refer to the redrawn Figure 7:

As there is no independent source present in the circuit which means voltage across capacitor is also $0\text{V}$.

So, value of $v_C\left(0^{-}\right)$ is $0\text{V}$.

At $t=0^{+}$ the capacitor does not allow sudden change in the voltage.

Therefore,

$v_C(0^{+})=v_C(0^{-})$

So,

$v_C\left(0^{+}\right)=0\text{ V}$

The circuit diagram is redrawn as shown in Figure 8for $t>0$.

At $t=\infty$ the capacitor in the $RC$ circuit is connected for long time.

So, capacitor behaves as open circuit.

The circuit diagram is redrawn as shown in Figure 9 for $t=\infty$.

Refer to the redrawn Figure 9:

The expression for KVL in mesh $FABEF$ is as follows:

$\left(i_1-i_2\right)R_1+v_1=0$ (21)

Here,

$i_1$ is current flowing in mesh $FABEF$,

$i_2$ is current flowing in mesh $FABEF$,

$R_1$ is resistance in the circuit and

$v_1$ is voltage in circuit.

Substitute $2\text{mA}$ for $i_1$, $10\text{Ω}$ for $R_1$ and $0.1v_x$ for $v_1$ in equation (21):

$\begin{array}{l}\left(2\text{mA}-i_2\right)\left(10\text{Ω}\right)+0.1v_x=0\\ \left(2\times {10}^{-3}\text{A}-i_2\right)\left(10\text{Ω}\right)+0.1v_x=0\left\{\because 1\text{mA}={10}^{-3}\text{A}\right\}\end{array}$

$0.02\text{V}-10i_2+0.1v_x=0$ (22)

The expression for voltage $v_x$ is as follows:

$v_x=i_2{R}_2$ (23)

Here,

$v_x$ is voltage across resistor $R_2$,

$i_2$ iscurrent in mesh $EBCDE$ and

$R_2$ is the resistance in the circuit.

Substitute $4\text{Ω}$ for $R_2$ in equation (23):

$v_x=i_2\left(4\text{Ω}\right)$

Substitute $4i_2$ in equation (22):

$\begin{array}{l}0.02\text{V}-10i_2+0.1\left(4i_2\right)=0\\ 0.02\text{V}-10i_2+0.4\left(i_2\right)=0\end{array}$

$0.02\text{V}-10.4i_2=0$ (24)

Rearrange equation (24) for $i_2$:

$10.4i_2=0.02$

Simplify for $i_2$,

$i_2=1.92\text{mA}$

Substitute $1.92\text{mA}$ for $i_2$ and $10\text{kΩ}$ for $R_3$ in equation (15):

$\begin{array}{c}v=\left(1.92\text{mA}\right)\left(10\text{kΩ}\right)\\ =\left(1.92\times {10}^{-3}\text{A}\right)\left(10\times {10}^3\text{Ω}\right)\left\{\begin{array}{l}\because 1\text{mA}={10}^{-3}\text{A}\\ \because 1\text{kΩ}={10}^3\text{Ω}\end{array}\right\}\\ =19.2\text{V}\end{array}$

So,

$v_C\left(\infty \right)=19.2\text{V}$

Substitute $19.2\text{V}$ for $v_C\left(\infty \right)$, $0\text{V}$ for $v_C(0^{+})$, $40.7446\text{ms}$ for $\tau$ in equation (16),

$v_C(t)=19.2\text{V}+\left(0\text{V}-19.2\text{V}\right)e^{\frac{-t}{40.7446\text{ms}}}$

$v_C(t)=19.2\left(1-e^{\frac{-t}{40.7446\text{ms}}}\right)\text{V}$ (25)

Substitute $19.2\left(1-e^{\frac{-t}{40.7446\text{ms}}}\right)\text{V}$ for $v_C$ and $\text{3 mF}$ for $C$ in equation (17):

$\begin{array}{c}{i}_C\left(t\right)=\text{3 mF}\frac{d}{dt}\left(19.2\left(1-e^{\frac{-t}{40.7446\text{ms}}}\right)\text{V}\text{V}\right)\\ =\text{3 mF}\left(19.2\right)\left(\frac{1}{40.7446\text{ms}}{e}^{\frac{-t}{40.7446\text{ms}}}\text{V}\right)\end{array}$

$i_C\left(t\right)=1.4137e^{\frac{-t}{40.7446\text{ms}}}\text{mA}$ (26)

Substitute $200\text{ms}$ for $t$ in equation (26):

$\begin{array}{c}{i}_C\left(200\text{ms}\right)=1.4137e^{\frac{-200\text{ms}}{40.7446\text{ms}}}\text{mA}\\ =0.01044\text{mA}\end{array}$

Substitute $200\text{ms}$ for $t$ in equation (25):

$\begin{array}{c}{v}_C\left(200\text{ms}\right)=19.2\left(1-e^{\frac{-200\text{ms}}{40.7446\text{ms}}}\right)\text{V}\\ =19.2\left(0.993\right)\text{V}\\ =19.06\text{V}\end{array}$

The expression for current $i_R$ is as follows:

$i_R\left(200\text{ms}\right)=\frac{v_C\left(200\text{ms}\right)}{R_3}$ (27)

Here,

$i_R$ is current flowing through resistor $R_3$,

$v_C\left(200\text{ms}\right)$ is the voltage across the capacitor and

$R_3$ is the resistance in the circuit.

Substitute $19.06\text{V}$ for $v_C\left(200\text{ms}\right)$ and $10\text{kΩ}$ for $R_3$ in equation (27):

$\begin{array}{c}{i}_R=\frac{19.06\text{V}}{10\text{kΩ}}\\ =\frac{19.06\text{V}}{10\times {10}^3\text{Ω}}\left\{\because 1\text{kΩ}={10}^3\text{Ω}\right\}\\ =0.001906\text{A}\end{array}$

The expression for current $i_{R_2}$ is as follows,

$i_{R_2}=i_R+i_C$ (28)

Here,

$i_R$ is current flowing through resistor $R_3$,

$i_C$ is capacitor current and

$i_{R_2}$ is current flowing through resistor $R_2$.

Substitute $0.001906\text{A}$ for $i_R\left(200\text{ms}\right)$ and $0.01044\text{mA}$ for $i_C\left(200\text{ms}\right)$ in equation (28):

$\begin{array}{c}{i}_{R_2}\left(200\text{ms}\right)=0.001906\text{A}+0.01044\text{mA}\\ =\text{0}\text{.01235}\text{A}\end{array}$

The expression for voltage $v_x$ is as follows:

$v_x=i_{R_2}{R}_2$ (29)

Here,

$v_x$ is voltage across resistor $R_2$,

$i_{R_2}$ is current flowing through resistor $R_2$ and

$R_2$ is the resistance in the circuit.

Substitute $\text{0}\text{.01235}\text{A}$ for $i_{R_2}\left(200\text{ms}\right)$ and $4\text{Ω}$ for $R_2$ in equation (29),

$\begin{array}{c}{v}_x=\left(\text{0}\text{.01235}\text{A}\right)4\text{Ω}\\ =0.04938\text{mV}\end{array}$

So value of voltage $v_x$ with dielectric is $0.04938\text{mV}$.

The expression for percentage change in voltage $v_x$ with dielectric resistor is as follows:

$\Delta v_x=\frac{v_{x_2}-v_x{}_{{}_1}}{v_x{}_{{}_2}}\times 100$ (14)

Here,

$\Delta v_x$ is percentage change in voltage $v_x$,

$v_x{}_{{}_1}$ isvalue of voltage $v_x$ without dielectric and

$v_{x_2}$ isvalue of voltage $v_x$ with dielectric.

Substitute $40.8\text{ms}$ for $v_x{}_{{}_1}$ and $40.7446\text{ms}$ for $v_{x_2}$ in equation (14):

$\begin{array}{c}\Delta v_x=\frac{0.04938\text{mV}-0.04372\text{mV}}{0.04938\text{mV}}\times 100\\ =\frac{0.00566\text{mV}}{0.04938\text{mV}}\times 100\\ =11.46\%\end{array}$

Conclusion:

Thus, value of voltage $v_x$ without dielectric is $0.04372\text{mV}$, value of voltage $v_x$ with dielectric is $0.04938\text{mV}$ and value of percentage change in voltage $v_x$ is $11.46\%$.