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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