Chapter 23.1, Problem 12PP

a.

To determine

Theequivalent resistance of the circuit.

Answer to Problem 12PP

The equivalent resistance in the circuit is $\text{11}\text{.9 k}\Omega$.

Explanation of Solution

Given:

Three resistors of value $3.3\text{ k}\Omega ,4.7\text{ k}\Omega \text{ and }3.9\text{ k}\Omega$ are connected in series across a 12-V battery.

Formula used:

For a set of resistors $R_1$ , $R_2$ , $R_3$ in series, the equivalent resistance can be calculated as,

  $R_{eq}=R_1+R_2+R_3+\mathrm{...}$ …… (1)

Calculation Consider the circuit shown in Figure 1. Here, the generator is supplying a voltage of 12-V and is loaded with three resistances in series, $3.3\text{ k}\Omega ,4.7\text{ k}\Omega \text{ and }3.9\text{ k}\Omega$

  

Figure 1

The equivalent resistance can be calculated using (1) as,

  $\begin{array}{c}{R}_{eq}=3.3+4.7+3.9\\ =11.9\text{ k}\Omega \end{array}$

Conclusion:

The equivalent resistance of the circuit is $11.9\text{ k}\Omega$ .

b.

To determine

The current in the circuit.

Answer to Problem 12PP

The current through the circuit is $1.01\text{ mA}$.

Explanation of Solution

Given:

Three resistors of value $3.3\text{ k}\Omega ,4.7\text{ k}\Omega \text{ and }3.9\text{ k}\Omega$ are connected in series across a 12-V battery.

Formula used:

For the resistance of the circuit and supply voltage, the current can be calculated using the Ohm’s law as,

  $\begin{array}{l}V=IR\\ I=\frac{V}{R}\end{array}$ …… (2)

Calculation:

The current flowing through the circuit can be calculated by using the relation in (2), by substituting the total resistance obtained in the part (a) as

  $\begin{array}{c}I=\frac{12}{11.9\times {10}^3}\\ =1.01\text{ mA}\end{array}$

Conclusion:

The current through the circuit is 1.01 mA.

c.

To determine

Thetotal potential difference across the three resistors.

Answer to Problem 12PP

The total potential difference across the three resistors is $12\text{ V}$

The potential differences across the three resistors are

  $V_{3.3\text{k}}=3.333\text{ V}$ , $V_{\text{4}\text{.7k}}=4.747\text{ V}$

  $V_{3.9\text{k}}=3.939\text{ V}$

Explanation of Solution

Given:

Three resistors of value $3.3\text{ k}\Omega ,4.7\text{ k}\Omega \text{ and }3.9\text{ k}\Omega$ are connected in series across a 12-V battery.

Formula used:

The total potential difference across the three resistors is the supply voltage itself. It can also be expressed as the product of the current through the circuit and the total resistance given by,

  $V=IR$

Calculation The total potential difference across the resistors can be expressed using (1) as,

  $\begin{array}{c}V=11.9\times {\text{10}}^3\times 1.01\times {\text{10}}^{-3}\\ =12\text{ V}\end{array}$

The individual potential difference across the resistors can be calculated using (1) as follows.

  $\begin{array}{c}{V}_{3.3\text{k}}=3.3\times {\text{10}}^3\times 1.01\times {\text{10}}^{-3}\\ =3.333\text{ V}\end{array}$

  $\begin{array}{c}{V}_{\text{4}\text{.7k}}=4.7\times {\text{10}}^3\times 1.01\times {\text{10}}^{-3}\\ =4.747\text{ V}\end{array}$

  $\begin{array}{c}{V}_{3.9\text{k}}=3.9\times {\text{10}}^3\times 1.01\times {\text{10}}^{-3}\\ =3.939\text{ V}\end{array}$

Conclusion:

Therefore, the total potential difference across the three resistors is $12\text{ V}$ .

The potential differences across the three resistors are

  $V_{3.3\text{k}}=3.333\text{ V}$ , $V_{\text{4}\text{.7k}}=4.747\text{ V}$

  $V_{3.9\text{k}}=3.939\text{ V}$