Chapter 27: Q36P (page 797)

In Figure, $ε_{1} = 6 .00 V$ , $ε_{2} = 12 .0 V$ , $R_{1} = 100 Ω$ , $R_{2} = 200 Ω$ ,and $R_{3} = 300 Ω$ . One point of the circuit is grounded $(V = 0)$ .(a)What is the size of the current through resistance 1? (b) What is the direction (up or down) of the current through resistance 1? (c) What is the size of the current through resistance 2?(d) What is the direction (left or right) of the current through resistance 2? (e) What is the size of the current through resistance 3? (f) What is the direction of the current through resistance 3? (g) What is the electric potential at point A?

Short Answer

Expert verified

(a) Magnitude of the current flowing through the resistance . $R_{1} = 0.0382 A$ .

(b)Direction of the current flowing through the resistance $R_{1}$ . is downward.

(d) Direction of the current flowing through the resistance $R_{2}$ is rightward.

(e) Magnitude of the current flowing through the resistance $R_{3} = 0.0273 A$ .

(f) Direction of the current flowing through the resistance $R_{3}$ is leftward.

(g) Electric potential at point Ais . $+ 3.82 V$

Step by step solution

Given data:

The emf voltage, $ε_{1} = 6.00 V$

The emf voltage, $ε_{2} = 12.0 V$

The resistor, $R_{1} = 100 Ω$

The resistor, $R_{2} = 200 Ω$

The resistor, $R_{3} = 300 Ω$

Understanding the concept:

Use the loop rule (Kirchhoff’s Voltage law) and the junction rule (Kirchhoff’s Current law) to find the current flowing through each resistance. Once you get the value of current,you can tell that in which direction it is moving.

Formula:

For any loop, $\sum V = 0$

At any junction $\sum I_{i} = \sum I_{o}$

The Ohm’s law is, $V = I R$

(a) Calculate the size of the current through resistance 1:

Using the junction rule,you can say that

$I_{1} = I_{2} + I_{3}$ ….. (1)

Using the loop rule,you can say that

$ε_{1} - I_{2} R_{2} - I_{1} R_{1} = 0$ ….. (2)

$ε_{2} - I_{3} R_{3} - I_{1} R_{1} = 0$ ….. (3)

Substituting the value of $I_{1}$ into equation (1).

$\begin{array}{l} ε_{1} - I_{2} R_{2} - (I_{2} + I_{3}) R_{1} = 0 \\ ε_{1} - (R_{2} + R_{1}) I_{2} - I_{3} R_{1} = 0 \\ (R_{2} + R_{1}) I_{2} + I_{3} R_{1} = ε_{1} \end{array}$

Substitute known values in the above equation.

$300 I_{2} + 100 I_{3} = 6.0$ ….. (4)

Now substitute equation (1) into equation (2).

$\begin{array}{l} ε_{2} - I_{3} R_{3} - (I_{2} + I_{3}) R_{1} = 0 \\ ε_{2} - I_{3} (R_{3} + R_{1}) - I_{2} R_{1} = 0 \\ I_{3} (R_{3} + R_{1}) + I_{2} R_{1} = ε_{2} \end{array}$

Substitute known values in the above equation.

$100 I_{2} + 400 I_{3} = 12.0$ ….. (5)

Solving equations (4) and (5), and multiply equation (5) with , you get,

$\begin{matrix} 300 I_{2} + 100 I_{3} - 300 I_{2} - 1200 I_{3} = 6.0 - 36.0 \\ - 1100 I_{3} = - 30.0 \\ I_{3} = 0.0273 A \end{matrix}$

Substitute the above value into equation (4).

$\begin{array}{l} 300 I_{2} + 100 (0.0273) = 6.0 \\ 300 I_{2} = 6.0 - 2.73 \\ I_{2} = 0.0109 A \end{array}$

Put $0.0109 A$ for $I_{2}$ and $0.0273 A$ for $I_{3}$ into equation (1).

$\begin{matrix} I_{1} = I_{2} + I_{3} \\ = (0.0109 + 0.0273) A \\ = 0.0382 A \end{matrix}$

(b) Calculate the direction (up or down) of the current through resistance 1:

Since the magnitude of the current flowing through the resistance $R_{1}$ is $0.0382 A$ , by using the diagram and the junction rule,you can say that the direction of the current flowing through the resistance $R_{1}$ is downward.