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Electric Current
Submitted by: I.D. 039622568
The problem:
Given the values: 61 = 1 V,62
R5 = E4;= 0.6 ‫ח‬, R ‫ = ד‬0.7 ‫ח‬
= 0.5 V,63 = 0.6 V,Rl = R2 = 0.5 ‫ח‬, R3 = 1 ‫ח‬, ~ = 0.4 ‫ח‬,
1. Calculate the current flowing through each resistor, and the potential difference between B
and A when the 8witch is open.
2. Calculate the current flowing through each resistor, and the potential difference between B
and C when the switch is closed.
A
82
~
Rs
14
The solution:
1. Since the switch is open we can disregard the middle branch of the circuit, leaving only a circuit
with resistors and voltage sources in series. By using Ohm's Law we find:
vt
61 + 63 = 1 + 0.6 = 1.6 V
Rt
R1
+ R2 +
R3
(1)
+ E4;+ R‫ = ד‬0.5 +
0.5 + 1 + 0.6 + 0.7 = 3.3
‫ח‬
~ =
61 + 63
= 1.6 = 0.485 A
Rt
Rl + R2 + R3 + E4;+ R ‫ד‬
3.3
With the current moving counter clockwise because of the direction of the voltage sources.
1
(2)
(3)
Now in order to calculate the potential difference between B and A, all we have to do is calculate
the voltage between these points:
VBA
= 61 - 1(Rl + R2) =
1-
1= 0.515 V
(4)
2. Now that the switch is closed we can no longer disregard the middle brance, and we have to use
Kirchhoff's laws.
We will choose theright node as our junction and assume that 11 comes from above, 12 from below
and 13 fiowsto the left .
Now, our first path will be clockwise through R1,61,R2,R3,62,R5,R4,
and our second will be
likewise clockwise through Rl, 61, R2' R3, R ‫ד‬, 63, E4;. giving us the following equations :
11 +12 = 13
(5)
=0
(6)
11Rl + 61 + I1Rz + I1R3 - I2R ‫ ד‬+ E:3- I2E4; = 0
(7)
11R1 + 61 + 11R2 + 11R3 + E:2+ 13R5 + 13~
1
Using simple math, we found the currents: 11 = -0.6 A ‫ ו‬12 = 0.3 A and 13 = -0.3 A, while the
negative sign signals that the direction in which these currents flow is opposite to the one we chose.
in order to find the potential difference between points B and C we calculate the voltage, keeping
in mind that 13 now has a new direction. thus :
Vnc = 13(R4 + R5) = 0.3(0.4 + 0.6) = 0.3 V
2
(8)
‫ז‬:_ b!"2
~
1- c? 7-)
Resistance
Submitted by: I.D. 065944332
The problem:
Find the effective resistance between the points 1 & 2:
1
2
The solution:
Currents:
11+ 14
1T = 12+15
11 = 13+12
15 = 14 +13
(1)
lT
(2)
(3)
(4)
Voltages:
V - 11R1 - 12R2 = 0
(5)
V - 15R5 - 14R4 = 0
(6)
V - 11R1 -13R3 - 15R5 = 0
(7)
RT = R1[(R ‫ן‬
In the case that R1
+ R2) ~ + (R2 + R3 + R4)R5] + R2[( ~ R5 + R3( ~ + R5 )]
R3(R2 + R4) + R5(R3 + R4) + R1(R2 + R3 + 2R5 )
= R2 = R4
= R one obtains 11
1
= 12 = 14= 15=
~1and 13=
(8)
o.
resistors cube
Submitted by: I.D. 066101528
The problem:
In the resistors cube you see in the sketch all resistors are equal and their resistance is R.
1. Find the equivalent resistance between points a and b.
2. How does the equivalent resistance changes if we connect point 4 with point 5 with a resistor
R?
5
a
b
3
The solution:
1. We can easi1y see that the cube is symmetrical for a current to junction 4 and 5 therefore the
potential is equal in those two points, same about points 3 and 6. If the potential is equal it means
that we can connect these two junctions into one and create the electrical circuit you see in sketch
1.
a---.----I
b--~-4
R
R
We simplify circuit 1 by connecting all the resistors in parallel into one resistor:
1
Rt
1
R
1
R
-=-+-=}R,.=-
R
2
a---;..~
b---!:-'
1
We simplify circuit 2 by connecting all the resistors in row into one resistor:
R
R
= - + R + - = 2R
Rt
2
2
now we have a circuit like in sketch 3.
a
b
~
45
RLd
~2R
2 YzR "3-‫ל‬, 6~
---J
We simplify circuit 3 by connecting all the resistors in parallel into one resistor :
1
Rt
2
2R
1
= R + 2R
Rt
=}
=5
now we have a circuit like in sketch 4.
.1
a----r-...l
2!5R
b---'-'
We simplify circuit 4 by connecting all the resistors in row into one resistor:
Rt
R
2R
R
‫ך‬R
2
5
2
5
=-+-+-=-
now we have a circuit like in sketch 5.
a
bI]
‫ןך‬5R
We simplify circuit 5 by connecting all resist ‫ס‬rs in parallel into ‫ס‬ne resistor :
1
1
Rt = R
‫ך‬R
5
+ ‫ך‬R
=}
Rt = 12
we found the equivalent resistance between p ‫ס‬ints a and b to be
‫ך‬R
Rt
=12
2. Of course, the equivalent resistance would not change - that is because the potential between
those two points is zero, therefore, there w‫ס‬uld be n ‫ ס‬current and the resistance would not change !
2
Electric current
Submitted by: I.D. 200475846
The problem:
Find the resistance between the bases of a circle conic with radii a and 2a, with a height L and
filled with material with a conductivity a.
The solution:
1. The equation we shall use to find the resistance is: dR = :sfi a' a is given to US, 80 to find the
total resistance we need to find the area S(x ).
In order to find the radius of the cone as a function of L we shall use basic geometry and we will
come up with the expression:
r(x)=a(L+x).
(1)
L
Now all we need to do is to substitute S(x ) = ‫ ד‬r2(x) into dR =
st:j(T '
2
dR =
L dx
a ‫ ד‬a2(L
(2)
+ x )2
Now we shall perform the integral. The integral boundaries are from 0 to L.
r
L
R
=
10
L2dx
a ‫ ד‬a2(L
+ x)2
L
=
(3)
2a ‫ ד‬a2
1
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