The current in the resistor \(R_2\) is \(\boxed{1\text{ A}}\).
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 \(6\) Ω.
The resistance of second resistor is \(8\) Ω.
The resistance of third resistor is \(2\) Ω.
The voltage of first cell is \(4\text{ V}\).
The voltage of second cell is \(14\text{ V}\).
Concept:
Kirchhoff’s law in left side loop:
\(i_1R_1+(i_1-i_2)R_2=\epsilon_1\)
Substitute \(6\) Ω for \(R_1\), \(8\) Ω for \(R_2\) and \(4\text{ V}\) for \(\epsilon_1\) in above equation and simplify.
\(7i_1-4i_2=2\) …… (I)
Kirchhoff’s law in right side loop:
\((i_2-i_1)R_2+i_2R_3=\epsilon_2\)
Substitute \(8\) Ω for \(R_2\), \(2\) Ω for \(R_3\) and \(14\text{ V}\) for \(\epsilon_2\) in above equation and simplify.
\(-4i_1+5i_2=7\) …… (II)
Multiply equation (I) with \(5\)
\(35i_1-20i_2=10\) …… (III)
Multiply equation (II) with \(4\)
\(-16i_1+20i_2=28\) …… (IV)
On adding equation (III) and equation (IV), we get
\(i_1=2\text{ A}\)
Substitute \(2\text{ A}\) for \(i_1\) in equation (I) and rearrange for \(i_2\) .
\(i_2=3\text{ A}\)
Current in second resistor:
\(\begin{aligned}I&=|i_1-i_2|\\&=1\text{ A}\end{aligned}\)
Thus, the current in the resistor \(R_2\) is \(\boxed{1\text{ A}}\).
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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.
