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Part A
What is the current through the 10Ω resistor in the figure (Figure 1)?
Part B
Is the current from left to right or right to left?

left to right

right to left
Relevant knowledge
Kirchhoff’s Current Law
Kirchhoff’s Current Law says that the total electric potential at a node (point) in an electrical circuit is zero. This means that the sum of all currents flowing into the node will equal the sum of currents flowing away.
Kirchhoff’s Voltage Law
The sum of all voltage drops in an open loop in an electrical circuit will equal the sum of all EMFs in the loop.
Kirchhoff’s law and the circuit diagram are the concepts that were used to answer the question.
First draw the circuit diagram. Next, show the current direction in each loop. Next, determine the current through each loop using Kirchhoff’s loop law.
Kirchhoff’s Voltage Law
This law says that the sum total of all possible differences in any closed circuit must be zero.
Therefore,
[katex]\sum {V = 0} [/katex]
Let us consider a circuit below
Two resistors are [katex]{R_1}{\rm{ and }}{R_2}[/katex] in the circuit.
To the circuit above, apply Kirchhoff’s voltage law.
[katex]\begin{array}{c}\\\varepsilon – I{R_1} – I{R_2} = 0\\\\\varepsilon = I\left( {{R_1} + {R_2}} \right)\\\end{array}[/katex]
(A)
Below is the circuit diagram.
Let [katex]{I_1}[/katex] be the current in both loops of the circuit.
To the circuit above, apply Kirchhoff’s law.
For the loop 1.
[katex]\begin{array}{c}\\12 + 5{I_1} + 3 + 10{I_1} – 10{I_2} = 0\\\\15 + 15{I_1} – 10{I_2} = 0\\\\15{I_1} = 10{I_2} – 15\\\end{array}[/katex]
[katex]{I_1} = \frac{{10{I_2} – 15}}{{15}}[/katex]
The loop 2.
[katex]9 + 5{I_2} + 10{I_2} – 10{I_1} – 3 = 0[/katex]
[katex]6 + 15{I_2} – 10{I_1} = 0[/katex]
Substitute the equation (1) for (2)
[katex]\begin{array}{c}\\6 + 15{I_2} – 10\left( {\frac{{10{I_2} – 15}}{{15}}} \right) = 0\\\\6 + 15{I_2} – 6.66{I_2} + 10 = 0\\\\8.334{I_2} = – 16\\\\{I_2} = – 1.92{\rm{ A}}\\\end{array}[/katex]
Substitute, [katex] – 1.92{\rm{ A}}[/katex]
[katex]{R_1}{\rm{ and }}{R_2}[/katex]0
The current through [katex]{R_1}{\rm{ and }}{R_2}[/katex]1
[katex]{R_1}{\rm{ and }}{R_2}[/katex]2
We chose the wrong direction for the current, from the positive terminal to the negative terminal.
The current through [katex]{R_1}{\rm{ and }}{R_2}[/katex]1
(B)
We chose the direction of the circuit from the positive to the negative terminal. The [katex]{R_1}{\rm{ and }}{R_2}[/katex]1 negative sign is in the calculated current.
The current direction in [katex]{R_1}{\rm{ and }}{R_2}[/katex]1Ans is Part A.
The current through the [katex]{R_1}{\rm{ and }}{R_2}[/katex]1Part A
The direction of current through [katex]{R_1}{\rm{ and }}{R_2}[/katex]1