Find the current in the 3.00 ω resistor. (note that three currents are given.)

uploaded image 26.23) 1.Find the current in the 3.00 rm Omega resistor. (Note that three currents are given.)

2.Find the unknown emfs {cal E}_1 and {cal E}_2.
3.Find the resistance R.

Answer

General guidance

Concepts and reason
The concepts required to solve the problem are Ohm’s law and Kirchhoff’s junction rule. First, from the given currents, find the current in the3.00Qresistor. Using Kirchhoff’s voltage law, find the unknown emfs. Similarly, using the Kirchhoff’s voltage law, find the unknown resistance.

Fundamentals

Ohm’s law states that current between two points in a circuit is proportional to the potential difference between the points. The constant of proportionality is called the resistance. V IR Here, is the potential difference, is the current andRis the resistance. Kirchhoff’s circuit laws for current and voltage are applied in electrical circuits. Kirchhoff’s current law states that the total current entering a nod is equal to the total current leaving the node. Kirchhoff’s voltage law states that the total voltage around a closed loop is zero.

Step-by-step

Step 1 of 4

(26.23.1) The below figure shows the direction of current flowing through the circuit. R
2.00A
ి
E1
h
f
b
4.00
3.002
6.00S
3.00AV
5.00A
с
e
d
కా The current in the linebcand in the linefecombine and flow together through the linehdas current. So, the current in the linehdis given by be
fe Here, beis the current in the linebcandis the current in the linefe. Substitute3.00 Aforbeand5.00 Aforto find the current. I = 3.00 A +5.00 A
=8.00 A

Part 26.23.1

The current in the 3.00Qis8.00 A


The current through theresistor is the sum of the current through theresistor and through theresistor.

Step 2 of 4

(26.23.2.1) Applying Kirchhoff’s voltage law in loopbcdhb, |(3.00 A) (4.00)1 (3.002)= e Substitute8.00Aforto find the emf. (3.00 A) (4.00)+(8.00 A) (3.002)= c, So, the emfis & 12V+24V
=36V

Part 26.23.2.1

The emfis36V


According to Kirchhoff’s second law, the voltage at a closed loop is zero. Applying this law in the loop, the emfis the sum of the products of the current and resistor in wire and the current and resistor in wire.

Step 3 of 4

(26.23.2.2) Applying Kirchhoff’s voltage law in loophdefh, |(5.00 A) (6.00s2)+1 (3.00 2)=E, Substitute8.00Aforto find the emf&2. (5.00 A) (6.00)+(8.00 A)(3.00) c,
= So, the emf&2is 30V 24V
E
=54V

Part 26.23.2.2

The emf&2is54 V


Applying Kirchhoff’s second law in the loop, the emfis the sum of the products of the current and resistor in wire and the current and resistor in wire.

Step 4 of 4

(26.23.3) Applying Kirchhoff’s voltage law in loopabfga, &2-6(
(2A)R
R=-&
2A Substitute36 Vforand 54 Vfor&2to find the resistanceR. 54V -36 V
R =
2A
18V
2A
=90

Part 26.23.3

The resistanceRis.


Applying Kirchhoff’s second law in the loop, the resistanceis the sum emfsand.

Answer

Part 26.23.1

The current in the 3.00Qis8.00 A

Part 26.23.2.1

The emfis36V

Part 26.23.2.2

The emf&2is54 V

Part 26.23.3

The resistanceRis.

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