Question

3. For the circuit shown in the figure, all quantities are accurate to 2 significant figures. What is the value of the current I₁? 

A. 0.32 A B 0.61 A C. 0.29 A D. 0.89 A E. 0.11A

Answer 1

Concepts and reason

The concepts used to solve this problem are Kirchhoff’s loop rule and Ohm’s law.

Initially, the equation for the left loop can be calculated using the Kirchhoff’s loop rule. Then, the equation for the right loop can be calculated using the Kirchhoff’s loop rule. Finally, the value of the current can be calculated by using the relation between the left and right loop equations.

Fundamentals

Kirchhoff’s laws are used to simplify complex circuits to find the unknown current voltage or resistances.

Kirchhoff’s loop rule states that the sum of the voltage differences around the loop is zero.

$\Sigma V = 0$

Ohm’s law states that the potential difference between two points is proportional to the current passing through the points. The proportionality constant is the resistance.

$V = IR$

Here, $V$ is the potential difference, $I$ is the current and $R$ is the resistance.

The expression for the equation for the right loop is given below:

$13{I_3} - 10{I_1} - {V_2} + {V_3} = 0$

Here, ${V_1}\;{\rm{and}}\;{V_2}$ are voltages connected in the circuit, ${I_1}$ is the current in the left loop, and ${I_3}$ is the current in the right loop.

Substitute $6\;{\rm{V}}$ for ${V_2}$ , $8\;{\rm{V}}$ for ${V_3}$

$\begin{array}{c}\\13{I_3} - 10{I_1} - \left( {6\;{\rm{V}}} \right) + \left( {8\;{\rm{V}}} \right) = 0\\\\{I_3} = \frac{{\left( {10{I_1}} \right) - \left( {2\;{\rm{V}}} \right)}}{{13}}\\\end{array}$

The expression for the current in the left loop is given below:

${V_1} + 15{I_1} + {V_2} - 10{I_3} = 0$

Here, ${V_3}$ is voltage connected in the circuit.

Substitute $\left( { - 12\;{\rm{V}}} \right)$ for ${V_1}$ , $6\;{\rm{V}}$ for ${V_2}$ , $\left[ {\frac{{\left( {10{I_1}} \right) - \left( {2\;{\rm{V}}} \right)}}{{13}}} \right]$ for ${I_3}$ .

$\begin{array}{c}\\\left( { - 12\;{\rm{V}}} \right) + 15{I_1} + \left( {6\;{\rm{V}}} \right) - 10\left( {\frac{{\left( {10{I_1}} \right) - \left( {2\;{\rm{V}}} \right)}}{{13}}} \right) = 0\\\\95{I_1} = 58\\\\{I_1} = 0.61\;{\rm{A}}\\\end{array}$

Ans:

The value of the current is ${\bf{0}}{\bf{.61}}\;{\bf{A}}$ .

Answer 2