Physics, 01.09.2019 18:20 fanofhope31

If r1 = 6 ω, r2 = 8 ω, r3 = 2 ω, ε1 = 4 v, and ε2 = 14 v, what is the current in r2?

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Answer from: netflixacc0107

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 Ω.

The resistance of second resistor is Ω.

The resistance of third resistor is 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 Ω for R_1 , 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 Ω for R_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

35i_1-20i_2=10 …… (III)

Multiply equation (II) with

-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 $\boxed{1\text{ A}}$ .

Learn more:

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2. Find the odd one out for lever:

3. Volume of gas after expansion:

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.

If r1 = 6 ω, r2 = 8 ω, r3 = 2 ω, ε1 = 4 v, and ε2 = 14 v, what is the current in r2?

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