Chapter 21, Problem 50PE

To determine

(a)

The current for the given circuit

Answer to Problem 50PE

  $I=1.5848\text{mA}$

Explanation of Solution

Given information:

The voltage of alkaline cell is $1.585$ V.

An internal resistance is $0.100\Omega$.

Voltmeter is used with $1k\Omega$

The given figure is

Introduction:

Kirchhoff's second rule states that in a closed loop the sum of all the voltages is equal to zero.

Kirchhoff's junction rule states that the sum of all the currents entering a junction is equal to the sum of all the currents leaving the junction.

Ohms law states that at constant temperature potential difference $V$ across the terminal is directly proportional to current $I$ flowing through the circuit

  $\text{V=IR}$

Where $R$ is a constant value called as resistance

Equivalent resistance of several resistors connected in series is

  ${\text{R}}_{\text{eq}}{\text{=R}}_{\text{1}}{\text{+R}}_{\text{2}}\text{+}\mathrm{............}{\text{+R}}_{\text{N}}$

It is known that

Ohms law states that at constant temperature potential difference $V$ across the terminal is directly proportional to current $V$ flowing through the circuit

  $\text{V=IR}$

Where $R$ is a constant value called as resistance

Equivalent resistance of several resistors connected in series is

  ${\text{R}}_{\text{eq}}{\text{=R}}_{\text{1}}{\text{+R}}_{\text{2}}\text{+}\mathrm{............}{\text{+R}}_{\text{N}}$

It is given that

  $V=1.585$ V

Resistors are connected in series so

  $R_{eq}=0.100+1000\left(1k\Omega =1000\Omega \right)=1000.1$

Substitute all the values in $V=IR$

  $\begin{array}{l}I=\frac{V}{R_{eq}}\\ \Rightarrow I=\frac{1.585}{1000.1}\\ \Rightarrow I=0.0015848\\ \Rightarrow I=1.5848\text{mA}\end{array}$

Conclusion:

The current for the given circuit is $I=1.5848\text{mA}$.

To determine

(b)

The terminal voltage of the given circuit

Answer to Problem 50PE

  $1.5848\text{V}$

Explanation of Solution

Given information:

The voltage of alkaline cell is $1.585$ V.

An internal resistance is $0.100\Omega$.

Voltmeter is used with $1k\Omega$

The given figure is

Introduction:

Kirchhoff's second rule states that in a closed loop the sum of all the voltages is equal to zero.

Kirchhoff's junction rule states that the sum of all the currents entering a junction is equal to the sum of all the currents leaving the junction.

Ohms law states that at constant temperature potential difference $V$ across the terminal is directly proportional to current $V$ flowing through the circuit

  $\text{V=IR}$

Where $R$ is a constant value called as resistance

Equivalent resistance of several resistors connected in series is

  ${\text{R}}_{\text{eq}}{\text{=R}}_{\text{1}}{\text{+R}}_{\text{2}}\text{+}\mathrm{............}{\text{+R}}_{\text{N}}$

It is known that

Ohms law states that at constant temperature potential difference $V$ across the terminal is directly proportional to current $V$ flowing through the circuit

  $\text{V=IR}$

Where $R$ is a constant value called as resistance

It is given that

  $I=1.5848\text{mA}$

  $R=$

  $1k\Omega$

Substitute all the values in $V=IR$

  $\begin{array}{l}V=IR\\ \Rightarrow V=1.5848\times 1=1.5848\text{V}\end{array}$

Conclusion:

The terminal voltage of the given circuit is $1.5848\text{V}$.

To determine

(c)

The ratio of the measured terminal voltage to the emf for the given circuit

Answer to Problem 50PE

  $0.9998$

Explanation of Solution

Given information:

The voltage of alkaline cell is $1.585$ V.

An internal resistance is $0.100\Omega$.

Voltmeter is used with $1k\Omega$

The given figure is

Introduction:

Kirchhoff's second rule states that in a closed loop the sum of all the voltages is equal to zero.

Kirchhoff's junction rule states that the sum of all the currents entering a junction is equal to the sum of all the currents leaving the junction.

Ohms law states that at constant temperature potential difference $V$ across the terminal is directly proportional to current $V$ flowing through the circuit

  $\text{V=IR}$

Where $R$ is a constant value called as resistance

Equivalent resistance of several resistors connected in series is

  ${\text{R}}_{\text{eq}}{\text{=R}}_{\text{1}}{\text{+R}}_{\text{2}}\text{+}\mathrm{............}{\text{+R}}_{\text{N}}$

The terminal voltage of the given circuit is $1.5848\text{V}$.

The given emf is $1.585$ V.

The ratio of the measured terminal voltage to the emf for the given circuit is

  $\frac{\text{measured terminal voltage}}{\text{given emf}}=\frac{1.5848}{1.585}=0.9998$

Conclusion:

The ratio of the measured terminal voltage to the emf for the given circuit is $0.9998$.