Chapter 18, Problem 92P

(a)

To determine

Find the initial energy stored in the capacitor.

Answer to Problem 92P

The initial energy stored in the capacitor is $\text{900 J}$.

Explanation of Solution

Write the equation for energy.

$U_0=\frac{1}{2}C{V}_0^2$ (I)

Here, $U_0$ is the initial energy, $C$ is the capacitance and $V_0$ is the initial voltage.

Conclusion:

Substitute $6.0\times {10}^3\text{ V}$ for $V_0$ and $\text{50}\text{.0}\times {\text{10}}^{-6}\text{ F}$ for $C$ in equation I.

$\begin{array}{c}{U}_0=\frac{1}{2}\left(\text{50}\text{.0}\times {\text{10}}^{-6}\text{ F}\right){\left(6.0\times {10}^3\text{ V}\right)}^2\\ =900\text{ J}\end{array}$

Therefore, the initial energy stored in the capacitor is $\text{900 J}$.

(b)

To determine

Find the initial current through the patient.

Answer to Problem 92P

The initial current through the patient is $\text{25 A}$.

Explanation of Solution

Write the equation for current,

$I_0=\frac{V_0}{R}$ (II)

Here, $R$ is the resistance, $V_0$ is the initial voltage and $I_0$ is the initial current

Conclusion:

Substitute $6.0\times {10}^3\text{ V}$ for $V_0$ and $240\Omega$ for $R$ in equation II.

$\begin{array}{c}{I}_0=\frac{6.0\times {10}^3\text{ V}}{240\Omega }\\ =25\text{ A}\end{array}$

Therefore, the initial current through the patient is $\text{25 A}$.

(c)

To determine

Find the energy dissipated in the patient during $1.0\text{ ms}$ of time.

Answer to Problem 92P

The energy dissipated in the patient during $1.0\text{ ms}$ of time is $\text{140 J}$.

Explanation of Solution

Write the equation for voltage with respect to time,

$V_C=V_0{e}^{-t/\tau }$ (III)

Here, $V_C$  is the voltage across the capacitor, $V_0$ is the initial voltage, $t$ is the time and $\tau$ is the time constant.

Write the equation for energy.

$U=\frac{1}{2}C{V}_C^2$ (IV)

Here, $U$ is the energy, $C$ is the capacitance and $V_C$ is the voltage across the capacitor.

Then, the energy dissipated in the patient is,

$U_0-U$

Substitute equation III in IV and then in the above equation.

$U_0-U=U_0-\frac{1}{2}C{\left(V_0{e}^{-t/\tau }\right)}^2=U_0\left(1-e^{-2t/RC}\right)$ (V)

Conclusion:

Substitute $\text{900 J}$ for $U_0$ , $0.0010\text{ s}$ for $t$ , $240\Omega$ for $R$ and $\text{50}\text{.0}\times {\text{10}}^{-6}\text{ F}$ for $C$ in equation V.

$\begin{array}{c}{U}_0-U=\left(\text{900 J}\right)\left(1-e^{-2\left(0.0010\text{ s}\right)/\left(240\Omega \right)\left(\text{50}\text{.0}\times {\text{10}}^{-6}\text{ F}\right)}\right)\\ =140\text{ J}\end{array}$

Therefore, the energy dissipated in the patient during $1.0\text{ ms}$ of time is $\text{140 J}$.

(d)

To determine

Compare the average power supplied by the power source with the power delivered by the patient.

Answer to Problem 92P

The average power supplied by power source is $0.0033$ times that delivered to the patient.

Explanation of Solution

Write the average power by source and patient the divide them.

$\begin{array}{c}\frac{P_{source}}{P_{patient}}=\frac{\frac{U_0}{\Delta t_{source}}}{\frac{U}{\Delta t_{patient}}}\\ =\frac{U_0\Delta t_{patient}}{U\Delta t_{source}}\end{array}$ (VI)

Conclusion:

Substitute $\text{900 J}$ for $U_0$ , $\text{138 J}$ for $U$, $1.0\times {10}^{-3}\text{ s}$ for $\Delta t_{patient}$ and $2.0\text{ s}$ for $\Delta t_{source}$ in equation VI.

$\begin{array}{c}\frac{P_{source}}{P_{patient}}=\frac{\left(\text{900 J}\right)\left(1.0\times {10}^{-3}\text{ s}\right)}{\left(\text{138 J}\right)\left(2.0\text{ s}\right)}\\ =0.0033\end{array}$

Therefore, the average power supplied by power source is $0.0033$ times that delivered to the patient.

(e)

To determine

Why the capacitors are used in the defibrillator.

Answer to Problem 92P

The capacitors can produce much higher burst of current to the patient than the power source.

Explanation of Solution

The capacitor used in the defibrillator can produces much higher burst of current to the patient when compared to the burst produces by the power source.

Conclusion:

Therefore, the capacitors are used to produce burst in current to the patients.