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```Electric Current
Submitted by: I.D. 039622568
The problem:
Given the values: ε1 = 1 V, ε2 = 0.5 V, ε3 = 0.6 V, R1 = R2 = 0.5 Ω, R3 = 1 Ω, R4 = 0.4 Ω,
R5 = R6 = 0.6 Ω, R7 = 0.7 Ω
1. Calculate the current flowing through each resistor, and the potential difference between B
and A when the switch is open.
2. Calculate the current flowing through each resistor, and the potential difference between B
and C when the switch is closed.
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 = ε1 + ε3 = 1 + 0.6 = 1.6 V
(1)
Rt = R1 + R2 + R3 + R6 + R7 = 0.5 + 0.5 + 1 + 0.6 + 0.7 = 3.3 Ω
ε1 + ε3
1.6
Vt
=
=
= 0.485 A
I =
Rt
R1 + R2 + R3 + R6 + R7
3.3
(2)
(3)
With the current moving counter clockwise because of the direction of the voltage sources.
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 = ε1 − I(R1 + R2 ) = 1 − I = 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 the right node as our junction and assume that I1 comes from above, I2 from below
and I3 flows to the left.
Now, our first path will be clockwise through R1 , ε1 , R2 , R3 , ε2 , R5 , R4 , and our second will be
likewise clockwise through R1 , ε1 , R2 , R3 , R7 , ε3 , R6 . giving us the following equations:
I1 + I2 = I3
(5)
I1 R1 + ε1 + I1 R2 + I1 R3 + ε2 + I3 R5 + I3 R4 = 0
(6)
I1 R1 + ε1 + I1 R2 + I1 R3 − I2 R7 + ε3 − I2 R6 = 0
(7)
1
Using simple math, we found the currents: I1 = −0.6 A, I2 = 0.3 A and I3 = −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 I3 now has a new direction. thus:
VBC = I3 (R4 + R5 ) = 0.3(0.4 + 0.6) = 0.3 V
2
(8)
‫א‪.‬בהתחלה ניתן לראות כי ברגע שהמפסק פתוח יש לנו מעגל טורי וניתן להשתמש בחןק אוהם למציאת הזרם על‬
‫הנגדים‬
‫‪1   2  I  R1  R2  R3  R4  R5   0‬‬
‫‪‬‬
‫‪I  0.578 A‬‬
‫הפרש פוטנציאלים ניתן למצוא כך‬
‫‪VAB  1  I  R1  R5   0.614V‬‬
‫ב‪ .‬מה שאנו צריכים לפתור ‪ 3‬משוואות עם ‪ 3‬נעלמים‬
‫‪I1  I 2  I 3‬‬
‫‪1   2  I1 ( R1  R2  R3 )  I 2 ( R4  R5 )  0‬‬
‫‪ 2   3  I 3 ( R6  R7 )  I 2 ( R4  R5 )  0‬‬
‫‪I1  0.636 A‬‬
‫‪I 2  0.285 A‬‬
‫‪I 3  0.351A‬‬
‫‪dU‬‬
‫אנו יודעים את הקשר בין ההספק לאנרגיה‪:‬‬
‫‪dt‬‬
‫לפי הציור‬
‫‪P‬‬
‫‪dU dEth‬‬
‫‪‬‬
‫ניתן לראות ש‪:‬‬
‫‪dt‬‬
‫‪dt‬‬
‫בשביל למצוא את השיפוע נבחר את הערכים של ‪ Eth‬ב‪( t=5 -‬מתוך נוחות) ונקבל‪:‬‬
‫‪40mJ‬‬
‫‪P1 ‬‬
‫‪ 8mW  0.008W‬‬
‫‪5 sec‬‬
‫‪20mJ‬‬
‫‪P2 ‬‬
‫‪ 4mW  0.004W‬‬
‫‪5 sec‬‬
‫כעת ניתן לפתור את השאלה בשני דרכים‪:‬‬
‫‪ P ‬וזה בעצם שווה לשיפוע של הגרף‪.‬‬
‫א‪ .‬נוכיח שחיבור ההספקים שווה להספק על הבטרייה‪:‬‬
‫‪P  I V  I (V1  V2 )  I V1  I 2V2  P1  P2  0.008W  0.004W  0.012W‬‬
‫‪2‬‬
‫ב‪ .‬נמצא את המתח על הבטרייה ונשתמש בנוסחה הידוע‬
‫‪0.008‬‬
‫‪P1  I 2V1  V1 ‬‬
‫‪ 0.002V‬‬
‫‪4‬‬
‫‪0.004‬‬
‫‪P2  I 2V2  V2 ‬‬
‫‪ 0.001V‬‬
‫‪4‬‬
‫ומכאן המתח על הסוללה הינו‪:‬‬
‫‪Vbat  V1  V2  0.003V‬‬
‫וההספק‪:‬‬
‫‪2‬‬
‫‪Pbat  I Vbat  4 * 0.003  0.012W‬‬
‫‪2‬‬
‫‪2‬‬
```