Chapter 19, Problem 19P

(a)

To determine

The effect on the voltage across each resistor when the switch is closed.

Answer to Problem 19P

The voltage across resistor $R_1$ remain same but the voltage across resistor $R_3$ and $R_4$ decreases.

Explanation of Solution

Introduction:

The current in the circuit varies with the resistance of the circuit. If the resistance increase then the current decreases.

The resistors of the resistance $R_1$ is in the series with the equivalent resistance of resistors $R_2,R_3$ and $R_4$ . Initially the switch is open, so the voltage drop takes place across equivalent resistance of the resistors $R_3$ and $R_4$ . The current through the resistor $R_1$ and parallel combination of the resistors $R_3$ and $R_4$ is same because these are connected in series connection.

When the switch become closed then the equivalent resistance of the parallel resistors $R_2,R_3$ and $R_4$ decreases due to addition of the resistor $R_2$ . The voltage drop across resistor $R_1$ remain same but the voltage drop across resistor $R_3$ and $R_4$ decreases.

Conclusion:

Therefore, the voltage across resistor $R_1$ remain same but the voltage across resistor $R_3$ and $R_4$ decreases.

(b)

To determine

The effect on the current through each resistor.

Answer to Problem 19P

The current through resistors $R_1$ and $R_2$ increases but the current through resistors $R_3$ and $R_4$ decreases due to closing of the switch.

Explanation of Solution

Introduction:

The current in the circuit varies with the resistance of the circuit. If the resistance decreases then the current increases.

The equivalent resistance of the circuit decreases due to closing of the switch, so the current drawn from the battery increases.

The resistor $R_1$ is in the series connection, so the current in the resistor $R_1$ increases. The switch is initially open, so initially current through the resistor $R_2$ is zero but current in the resistor $R_2$ increases due to closing of the switch.

The equivalent resistance of the parallel connection decreases due to addition of the resistor in parallel connection with closing of the switch. The current through resistors $R_3$ and $R_4$ decreases due to closing of the switch.

Conclusion:

Therefore, the current through resistors $R_1$ and $R_2$ increases but the current through resistors $R_3$ and $R_4$ decreases due to closing of the switch.

(c)

To determine

The effect on the power output of the battery due to closing of switch.

Answer to Problem 19P

The power output of the battery increase due to closing of switch.

Explanation of Solution

Introduction:

The power of the circuit varies with the resistance of the circuit. If the resistance decreases then the power of the circuit increases.

The equivalent resistance of the circuit with open switch is greater than the equivalent resistance with closed switch. The equivalent resistance of the circuit becomes less on closing of switch due to addition of the resistor $R_2$ in the circuit, so the power output of the battery increase due to closing of switch.

Conclusion:

Therefore, the power output of the battery increase due to closing of switch.

(d)

To determine

The current through each resistor before and after closing the switch.

Answer to Problem 19P

The current through resistor $R_1$ before and after closing of switch is $0.12\text{A}$ and $0.132\text{A}$ , the current through resistor $R_{3/4}$ and $R_{2/3/4}$ before and after closing of switch is $0.6\text{A}$ .

Explanation of Solution

Given info:

Resistance of the each resistor, $R_1=R_2=R_3=R_4=125\Omega$

Voltage of the battery in the circuit is, $V=22\text{V}$

Formula Used:

The expression to calculate equivalent resistance before close switch is,

  $R_{e1}=R_1+\frac{R_{3/4}}{2}$

The expression to calculate equivalent resistance after close switch is,

  $R_{e2}=R_1+\frac{R_{2/3/4}}{2}$

The expression to calculate current through resistor $R_1$ before close switch is,

  $I_i=\frac{V}{R_{e1}}$

The expression to calculate current through resistor $R_1$ after close switch is,

  $I_f=\frac{V}{R_{e2}}$

The expression to calculate current through resistors $R_3$ and $R_4$ before close switch is,

  $I_{3/4}=\frac{I_i}{2}$

The expression to calculate current through resistor $R_3$ and $R_4$ after close switch is,

  $I_{2/3/4}=\frac{I_f}{3}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{R}_{e1}=125\Omega +\frac{125\Omega }{2}\\ =187.5\Omega \end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{R}_{e2}=125\Omega +\frac{125\Omega }{3}\\ =166.67\Omega \end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{I}_i=\frac{22\text{V}}{187.5\Omega }\\ =0.12\text{A}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{I}_f=\frac{22\text{V}}{166.67\Omega }\\ =0.132\text{A}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{I}_{3/4}=\frac{0.12\text{A}}{2}\\ =0.6\text{A}\end{array}$

Substitute all the values in the above expression.

  $\begin{array}{c}{I}_{2/3/4}=\frac{0.132\text{A}}{3}\\ =0.044\text{A}\end{array}$

Conclusion:

Therefore, the current through resistor $R_1$ before and after closing of switch is $0.12\text{A}$ and $0.132\text{A}$ , the current through resistor $R_{3/4}$ and $R_{2/3/4}$ before and after closing of switch is $0.6\text{A}$ and $0.044\text{A}$ .