Chapter 19, Problem 1P

(a)

To determine

The terminal voltage for a batter if the battery is connected in series with an $81\Omega$ resistor.

Answer to Problem 1P

The terminal voltage for a battery with an $81\Omega$ resistor in series is $8.40\text{V}$ .

Explanation of Solution

Given information:

Internal resistance of battery, $r=0.900\Omega$

EMF of the battery, $\epsilon =8.50\text{V}$

Resistance connected in series, $R_1=81\Omega$

Formula used:

Write the expression to calculate the current in the circuit.

  $V=\epsilon -Ir$............. (1)

Where,

• The current in the circuit is $I$ .

Write the expression to calculate the voltage across resistor $R_1$ using Ohm’s law.

  $V=I{R}_1$............. (2)

Where,

• The terminal voltage is $V$ .

Calculation:

Draw the circuit diagram.

  

Substitute $I{R}_1$ for $V$ in equation (1) from equation (2).

  $\begin{array}{c}I{R}_1=\epsilon -Ir\\ I{R}_1+Ir=\epsilon \\ I\left(R_1+r\right)=\epsilon \\ I=\frac{\epsilon }{R_1+r}\end{array}$

Substitute the values in the above expression.

  $\begin{array}{c}I=\frac{8.50}{81+0.900}\\ =\frac{8.50}{81.9}\\ =0.103\text{A}\end{array}$

Substitute the value of the current in equation (1).

  $\begin{array}{c}V=8.50-\left(0.103\right)\times 0.900\\ =8.50-0.0927\\ =8.40\text{V}\end{array}$

Therefore, the terminal voltage for a battery with an $81\Omega$ resistor in series is $8.40\text{V}$ .

(b)

To determine

The terminal voltage for a batter if the battery is connected in series with an $810\Omega$ resistor.

Answer to Problem 1P

The terminal voltage for a battery with an $810\Omega$ resistor in series is $8.49\text{V}$ .

Explanation of Solution

Given information:

Internal resistance of battery, $r=0.900\Omega$

EMF of the battery, $\epsilon =8.50\text{V}$

Resistance connected in series, $R_2=810\Omega$

Formula used:

Write the expression to calculate the current in the circuit.

  $V=\epsilon -Ir$............. (1)

Where,

• The current in the circuit is $I$ .

Write the expression to calculate the voltage across resistor $R_1$ using Ohm’s law.

  $V=I{R}_2$............. (3)

Where,

• The terminal voltage is $V$ .

Calculation:

Draw the circuit diagram.

  

Substitute $I{R}_2$ for $V$ in equation (1) from equation (3).

  $\begin{array}{c}I{R}_2=\epsilon -Ir\\ I{R}_2+Ir=\epsilon \\ I\left(R_2+r\right)=\epsilon \\ I=\frac{\epsilon }{R_2+r}\end{array}$

Substitute the values in the above expression.

  $\begin{array}{c}I=\frac{8.50}{810+0.900}\\ =\frac{8.50}{810.9}\\ =0.010\text{A}\end{array}$

Substitute the value of the current in equation (1).

  $\begin{array}{c}V=8.50-\left(\mathrm{0.0.10}\right)\times 0.900\\ =8.50-0.0927\\ =8.49\text{V}\end{array}$

Therefore, the terminal voltage for a battery with an $810\Omega$ resistor in series is $8.49\text{V}$ .