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26.23) 1.Find the current in the 3.00 resistor. (Note that three currents are given.) 2.Find the...

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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.

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Answer #1
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 the3.00Q
resistor. 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 andR
is 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.

(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 linebc
and in the linefe
combine and flow together through the linehd
as current.

So, the current in the linehd
is given by

be
fe

Here, be
is the current in the linebc
andis the current in the linefe
.

Substitute3.00 A
forbe
and5.00 A
forto find the current.

I = 3.00 A +5.00 A
=8.00 A

(26.23.2.1)

Applying Kirchhoff’s voltage law in loopbcdhb
,

|(3.00 A) (4.00)1 (3.002)= e

Substitute8.00A
forto find the emf.

(3.00 A) (4.00)+(8.00 A) (3.002)= c,

So, the emfis

& 12V+24V
=36V

(26.23.2.2)

Applying Kirchhoff’s voltage law in loophdefh
,

|(5.00 A) (6.00s2)+1 (3.00 2)=E,

Substituteforto find the emf&2
.

(5.00 A) (6.00)+(8.00 A)(3.00) c,
=

So, the emf&2
is

30V 24V
E
=54V

(26.23.3)

Applying Kirchhoff’s voltage law in loopabfga
,

&2-6(
(2A)R
R=-&
2A

Substitute36 V
forand 54 V
for&2
to find the resistanceR
.

54V -36 V
R =
2A
18V
2A
=90

Ans: Part 26.23.1

The current in the 3.00Q
is8.00 A

Part 26.23.2.1

The emfis36V

Part 26.23.2.2

The emf&2
is54 V

Part 26.23.3

The resistanceR
is.

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