Question

1. The ammeter in the figure reads 3.0 A.

a) Find I₂. Express your answer using two significant figures, in Amps.

b) Find ε. Express your answer using two significant figures, in Volts.

Answer 1

Concepts and reason

The concept used here is Kirchhoff’s current law and Kirchhoff’s voltage law.

Firstly, apply Kirchhoff’s current law to the left loop and Kirchhoff’s voltage law to the right loop and use the equation obtained from the loop to calculate the current ${I_2}$ .

In the second part, calculate the value of $\varepsilon$ by using the value of current calculated in the previous part.

Fundamentals

Kirchhoff’s current law:

At any junction in circuit, the algebraic sum of currents will be zero.

Kirchhoff’s voltage law:

In any closed loop of circuit, the algebraic sum of voltages will be zero.

The following figure shows the given circuit diagram.

(a)

The Kirchhoff’s current law is used at junction A, it is written as,

${I_1} + {I_2} = {I_{AB}}$ …… (1)

The current between A and B, it means the current ${I_{AB}}$ will be equal to the reading of attached ammeter, which is ${\rm{3 A}}$ . Therefore, it is written as,

${I_{AB}} = {\rm{3 A}}$

The voltage difference between A and B is written as,

${V_{AB}} = {I_{AB}}{R_{AB}}$

Substitute ${\rm{3 A}}$ for ${I_{AB}}$ and ${\rm{2 }}\Omega$ for ${R_{AB}}$ in above expression.

$\begin{array}{c}\\{V_{AB}} = \left( {{\rm{3 A}}} \right)\left( {{\rm{2 }}\Omega } \right)\\\\ = {\rm{6 V}}\\\end{array}$

Apply the Kirchhoff’s voltage law in loop 1, it is written as,

$9{\rm{ V}} - 3{I_1} = {V_{AB}}$

Substitute the value ${\rm{6 V}}$ for ${V_{AB}}$ in above expression.

$\begin{array}{c}\\9{\rm{ V}} - 3{I_1} = 6{\rm{ V}}\\\\3{I_1} = {\rm{3 V}}\\\\{I_1} = {\rm{1 A}}\\\end{array}$

Substitute the value ${\rm{3 A}}$ for ${I_{AB}}$ and ${\rm{1 A}}$ for ${I_1}$ in expression (1).

$\begin{array}{c}\\{\rm{1 A}} + {I_2} = {\rm{3 A}}\\\\{I_2} = {\rm{2 A}}\\\end{array}$

(b)

Apply the Kirchhoff’s voltage law in loop 2, it is written as,

$\varepsilon - 4.5{I_2} = {V_{AB}}$

Substitute the value ${\rm{6 V}}$ for ${V_{AB}}$ and ${\rm{2 A}}$ for ${I_2}$ in above expression.

$\begin{array}{c}\\\varepsilon - 4.5\left( {{\rm{2 A}}} \right) = {\rm{6 V}}\\\\\varepsilon = 1{\rm{5 V}}\\\end{array}$

Ans: Part a

The value of ${I_2}$ is ${\rm{2 A}}$ .

Part b

The value of $\varepsilon$ is $1{\rm{5 V}}$ .