The current in the resistor [tex]R_2[/tex] is [tex]\boxed{1\text{ A}}[/tex].
Further Explanation:
There are two rectangular loops in the given circuit.
To obtain the current in both the loops we will apply Kirchhoff’s voltage law.
Given:
The resistance of first resistor is [tex]6[/tex] Ω.
The resistance of second resistor is [tex]8[/tex] Ω.
The resistance of third resistor is [tex]2[/tex] Ω.
The voltage of first cell is [tex]4\text{ V}[/tex].
The voltage of second cell is [tex]14\text{ V}[/tex].
Concept:
Kirchhoff’s law in left side loop:
[tex]i_1R_1+(i_1-i_2)R_2=\epsilon_1[/tex]
Substitute [tex]6[/tex] Ω for [tex]R_1[/tex], [tex]8[/tex] Ω for [tex]R_2[/tex] and [tex]4\text{ V}[/tex] for [tex]\epsilon_1[/tex] in above equation and simplify.
[tex]7i_1-4i_2=2[/tex] …… (I)
Kirchhoff’s law in right side loop:
[tex](i_2-i_1)R_2+i_2R_3=\epsilon_2[/tex]
Substitute [tex]8[/tex] Ω for [tex]R_2[/tex], [tex]2[/tex] Ω for [tex]R_3[/tex] and [tex]14\text{ V}[/tex] for [tex]\epsilon_2[/tex] in above equation and simplify.
[tex]-4i_1+5i_2=7[/tex] …… (II)
Multiply equation (I) with [tex]5[/tex]
[tex]35i_1-20i_2=10[/tex] …… (III)
Multiply equation (II) with [tex]4[/tex]
[tex]-16i_1+20i_2=28[/tex] …… (IV)
On adding equation (III) and equation (IV), we get
[tex]i_1=2\text{ A}[/tex]
Substitute [tex]2\text{ A}[/tex] for [tex]i_1[/tex] in equation (I) and rearrange for [tex]i_2[/tex] .
[tex]i_2=3\text{ A}[/tex]
Current in second resistor:
[tex]\begin{aligned}I&=|i_1-i_2|\\&=1\text{ A}\end{aligned}[/tex]
Thus, the current in the resistor [tex]R_2[/tex] is [tex]\boxed{1\text{ A}}[/tex].
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Answer Details:
Grade: College
Subject: Physics
Chapter: Electricity
Keywords:
R1, R2, R3, E1, E2, current, Kirchhoff’s, voltage, loop, circuit, resistor, 6 ohm, 2 ohm, 8 ohm, 4 volt, 14 volt and law.
