Q39CQ An electronic apparatus may have... [FREE SOLUTION]

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Chapter 21: Q39CQ (page 773)

An electronic apparatus may have large capacitors at high voltage in the power supply section, presenting a shock hazard even when the apparatus is switched off. A "bleeder resistor" is therefore placed across such a capacitor, as shown schematically in Figure 21.50, to bleed the charge from it after the apparatus is off. Why must the bleeder resistance be much greater than the effective resistance of the rest of the circuit? How does this affect the time constant for discharging the capacitor?

Short Answer

Expert verified

The rest of the circuit will use less power.

Increase the external resistor's resistance to increase the time constant.

Step by step solution

Concept of the problem

An electronic apparatus has a large capacitor to accommodate a large amount of charge depending upon the requirement of the apparatus, even when the apparatus is switched off the capacitor present in the apparatus stores the charge and when touched directly presents a shock hazard. To avoid the shock, a bleeder resistor is placed across such a capacitor to take out the charge which is stored in the capacitor when the apparatus is switched off.

Explanation

In a parallel circuit, the higher the resistance, the more voltage is distributed evenly across all branches. The sum of all branches equals the current.

This means that each branch's power can be estimated as follows:

\({\rm{P}} = {\rm{V}} \times {\rm{I}}\)

where \({\rm{V}}\) is the same for all branches, meaning need to calculate how much \({\rm{I}}\) is in each branch so could compare power relations between branches.

Again,

\({\rm{I}} = \frac{{\rm{V}}}{{\rm{R}}}\)

This means that the higher the resistance, the less current, and thus power, flows through that specific branch. This also indicates that the higher the resistor's resistance, the less electricity the rest of the circuit will consume. However, increasing its resistance will raise the time constant.