Chapter 21, Problem 1CQ

1. A switch has a variable resistance that is nearly zero when closed and extremely large when open, and it is placed in series with the device it controls. Explain the effect the switch in Figure 21.43 has on current when open and when closed.

To determine

The effect of switch on current when open and closed.

Answer to Problem 1CQ

Current increases from zero to some constant value as the switch is closed from initial open state.

Explanation of Solution

Given:

Resistance of the switch when closed $=R_{closed}=0\text{ ohm}$

Resistance of switch when open $=R_{open}=\infty \text{ ohm}$

Emf of the battery $=E$

Internal resistance of the battery $=r$

External resistance connected in series with battery $=R$

Current in the circuit when switch is closed $=i_{closed}$

Current in the circuit when switch is open $=i_{open}$

Formula Used:

Equivalent resistance in series is given as

$R_{eq}=R_1+R_2+R_3+\mathrm{....}$

According to ohm's law

  $i=\frac{V}{R}$

Calculation:

When the switch is open:

Equivalent resistance in series of the circuit given as

$R_{eq}=R+r+R_{open}$

Using ohm's law, current in the circuit is given as

  $\begin{array}{l}{i}_{open}=\frac{E}{R_{eq}}\\ i_{open}=\frac{E}{R+r+R_{open}}\end{array}$

Since, $R_{open}=\infty$

  $\begin{array}{l}{i}_{open}=\frac{E}{R+r+\infty }\\ i_{open}=\frac{E}{\infty }=0\text{ A}\end{array}$

When the switch is closed:

Equivalent resistance in series of the circuit given as

$R_{eq}=R+r+R_{closed}$

Using ohm's law, current in the circuit is given as

  $\begin{array}{l}{i}_{open}=\frac{E}{R_{eq}}\\ i_{open}=\frac{E}{R+r+R_{closed}}\end{array}$

Since $R_{closed}=0$

  $\begin{array}{l}{i}_{closed}=\frac{E}{R+r+0}\\ i_{closed}=\frac{E}{R+r}\end{array}$

Clearly $i_{closed}>i_{open}$

Conclusion:

When switch is closed, there is some constant value of current in the circuit and when the circuit is open, there is no current in the circuit.