Chapter 29, Problem 29P

(a)

To determine

The average current.

Answer to Problem 29P

8.1 mA

Explanation of Solution

Given:

Voltage, $120V_{rms}$

Resistance, $R=21k\Omega$

Capacitance, $C=25\mu f=25\times {10}^{-6}F$

Formula used:

It is known that $\text{V=IR}$

Where,

  $I$ is current.

  $V$ is voltage

  $R$ is resistance

Calculation:

First calculate the time constant for the circuit:

  $\text{τ}\text{=}\text{RC}\text{=}\left(\text{21}\times {\text{10}}^{\text{3}}\text{Ω}\right)\left({\text{25×10}}^{\text{-6}}\text{}\text{F}\right)\text{=}\text{0}\text{.53}\text{s}$

Because there are two peaks per cycle, the period of the rectified voltage is

  $\text{T = }\frac{\text{1}}{\text{2f}}\text{ = }\frac{\text{1}}{\text{ 2}\left(\text{60}\text{Hz}\right)}\text{ = 0}\text{.0083}\text{s}$

Because $\tau >>T$ the voltage across the capacitor will be essentially constant during a cycle.

Therefore, voltage will be the peak voltage.

So, the average current is

  ${\text{I}}_{\text{av}}\text{=}\frac{{\text{V}}_{\text{0}}}{\text{R}}\text{=}\frac{\sqrt{2}\left(120\text{V}\right)}{\left(21\text{kΩ}\right)}\text{=}\text{8}\text{.1}\text{mA}$

(b)

To determine

The average current for the given value of C.

Answer to Problem 29P

  $\text{5}\text{.7}\text{mA}\left(\text{rippled}\right)$ .

Explanation of Solution

Given:

Voltage, $120V_{rms}$

Resistance, $R=21k\Omega$

Capacitance, $C=0.10\mu f=\text{0}{\text{.10×10}}^{\text{-6}}\text{}\text{F}$

Formula used:

It is known that $\text{V=IR}$

Where,

  $I$ is current.

  $V$ is voltage

  $R$ is resistance

Calculation:

First calculate the time constant for the circuit:

  $\text{τ}\text{=}\text{RC}\text{=}\left(\text{21}\times {\text{10}}^{\text{3}}\text{Ω}\right)\left(\text{0}{\text{.10×10}}^{\text{-6}}\text{}\text{F}\right)\text{=}\text{0}\text{.0021}\text{s}$

Because $\tau <T$ the voltage across the capacitor will be ripped.

Therefore, average voltage will be close to rms voltage.

So, the average current is

  ${\text{I}}_{\text{av}}\text{=}\frac{{\text{V}}_{\text{rms}}}{\text{R}}\text{=}\frac{\left(\text{120}\text{V}\right)}{\left(\text{21}\text{kΩ}\right)}\text{=}\text{5}\text{.7}\text{mA}\left(\text{rippled}\right)$