Chapter 19, Problem 82GP

(a)

To determine

The time when the light flash.

Answer to Problem 82GP

The time when the light flash is 0.686 s.

Explanation of Solution

Given information:

Capacitance, $C=0.105\mu \text{F}$

Resistance, $R=2.35\times {10}^6\Omega$

Voltage, $V_0=90\text{V}$

Emf, $\epsilon =\text{105 V}$

Formula Used:

The formula is given by,

  $V_0=\epsilon \left(1-e^{-t/RC}\right)$

Here, $V_0$ is the voltage, $\epsilon$ is the emf, R and C is the resistance and the capacitance.

Calculation:

Solve the above equation.

  $t=-RC\ln \left(1-\frac{V_0}{\epsilon }\right)$

Substitute the values in the above equation.

  $\begin{array}{c}t=-RC\ln \left(1-\frac{V_0}{\epsilon }\right)\\ =-2.35\times {10}^6\Omega \left(0.105\mu \text{F}\right)\left(1-\frac{90\text{V}}{\text{105 V}}\right)\\ =0.686\text{ s}\end{array}$

Conclusion:

Thus, the time when the light flash is 0.686 s.

(b)

To determine

The time when value of R is increased.

Answer to Problem 82GP

The time will also increase when value of R is increased.

Explanation of Solution

Given information:

Capacitance, $C=0.105\mu \text{F}$

Resistance, $R=2.35\times {10}^6\Omega$

Voltage, $V_0=90\text{V}$

Emf, $\epsilon =\text{105 V}$

Formula Used:

The formula is given by,

  $V_0=\epsilon \left(1-e^{-t/RC}\right)$

Here, $V_0$ is the voltage, $\epsilon$ is the emf, R and C is the resistance and the capacitance.

Calculation:

Refer the above equation.

  $t=-RC\ln \left(1-\frac{V_0}{\epsilon }\right)$

The R and t are directly proportional to each other, so time will also increase when value of R is increased.

Conclusion:

Thus, the time will also increase when value of R is increased.

(c)

To determine

The reason for the brief flashing.

Answer to Problem 82GP

The reason for the brief flashing is that the neon lamp acts like a wire.

Explanation of Solution

Given information:

Capacitance, $C=0.105\mu \text{F}$

Resistance, $R=2.35\times {10}^6\Omega$

Voltage, $V_0=90\text{V}$

Emf, $\epsilon =\text{105 V}$

Formula Used:

The formula is given by,

  $V_0=\epsilon \left(1-e^{-t/RC}\right)$

Here, $V_0$ is the voltage, $\epsilon$ is the emf, R and C is the resistance and the capacitance.

Calculation:

When the threshold voltage is reached, the neon lamp acts like a wire with the minimal resistance.

Conclusion:

Thus, the reason for the brief flashing is that the neon lamp acts like a wire.

(d)

To determine

The condition after the lamp flashed for the first time.

Answer to Problem 82GP

The flash time is 0.686 s which is the time taken by the capacitor to discharge the capacitor.

Explanation of Solution

Given information:

Capacitance, $C=0.105\mu \text{F}$

Resistance, $R=2.35\times {10}^6\Omega$

Voltage, $V_0=90\text{V}$

Emf, $\epsilon =\text{105 V}$

Formula Used:

The formula is given by,

  $V_0=\epsilon \left(1-e^{-t/RC}\right)$

Here, $V_0$ is the voltage, $\epsilon$ is the emf, R and C is the resistance and the capacitance.

Calculation:

The flash time is 0.686 s which is the time taken by the capacitor to discharge the capacitor and then charge to the threshold voltage.

Conclusion:

Thus, the flash time is 0.686 s which is the time taken by the capacitor to discharge the capacitor.