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Capacitance
Submitted by: I.D. 200475846
The problem:
For the following electric circuit find
1. effective capacitance of the capacitors
2. energy of the effective capacitor
3. potentials and charges on C1 , C3 , C4
The solution:
1. First, we will look at the circuit and notice that the capacitor C5 is shorted.
Thus, we can ignore this capacitor when we calculate the effective capacitance.
We use ⊕ for the connection in serious and || for the connection in parallel.
Ca = C2 ⊕ C2 = 4 ⊕ 4 = 2µF
(1)
Cb = Ca ||C3 = 2 + 1.5 = 3.5µF
(2)
Cc = Cb ||C4 = 3.5||5 = 8.5µF
(3)
Ceff = Cc ⊕ C1 = 8.5 ⊕ 3 = 2.22µF
(4)
2. The energy is
2.11 · 10−6 · 202
1
U = CV 2 =
= 4.42 · 10−4 J
2
2
3. Using the different loops in the circuit we write:
q4
q1
V =
+
C1 C4
q1
q3
V =
+
C1 C3
q1
q2
V =
+2
C1
C2
q1 = q2 + q3 + q4
(5)
(6)
(7)
(8)
(9)
After solving the equations we get
q1 = 4.43 × 10−5 C
(10)
−5
C
(11)
−5
C
(12)
−5
C
(13)
q2 = 1.10 × 10
q3 = 0.78 × 10
q4 = 2.61 × 10
1
Electric dipole and capacitors
Submitted by: I.D. 039409040
The problem:
In the circuit shown on the work page the capacitors are not charged at the beginning. The switch
S is turned left and the first capacitor is charged. The switch is turned right. What are the charges
on all the capacitors?
The solution:
The capacitors C2 , C3 are connected in a series so their charges will be equal q2 = q3 . Let us replace
C3
them with an equivalent capacitor C23 = CC22+C
with the same charge.
3
When S was turned left C1 was charged with q = C1 V0 . The total charge is conserved after the
switch is turned right. We also know that V is equal on C1 and C23 . Let us write the following
equations:
q = q1 + q2 = q1 + q3
q2
q1
V23 = V1 =
=
C23
C1
1
q1
1
+
=
q2
C2 C3
C1
q1 = q − q2 = C1 V0 − q2
1
1
q2
q2
+
= V0 −
C2 C3
C1
1
1
1
+
+
= V0
q2
C1 C2 C3
V0
q2 = q3 = 1
1
1
+
C1
C2 + C3
q1 = C1 V0 − q2
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
1
‫‪RC 5305‬‬
‫נתון המעגל הבא‪:‬‬
‫אין תלות בין הסעיפים‪.‬‬
‫א‪ .‬בזמן ‪ t = 0‬הקבל אינו טעון והמתג נסגר‪ .‬מהו הזרם ההתחלתי דרך הקבל בכיוון השעון‬
‫בזמן ‪?t = 0‬‬
‫ב‪ .‬המתג סגור זמן רב‪ ,‬לאחר מכן‪ ,‬בזמן ‪ ,t = 0‬המתג נפתח‪ .‬מהו הזרם ההתחלתי ימינה‬
‫בזמן ‪) ?t = 0‬כיוון חיובי הינו עם כיוון השעון(‬
‫פתרון‪:‬‬
‫א‪ .‬חוק זרמים‪:‬‬
‫‪I1 = I2 + Ic‬‬
‫חוק המתחים‪:‬‬
‫)‬
‫‪Vc‬‬
‫‪+ Ic R1‬‬
‫‪R2‬‬
‫‪Vc‬‬
‫‪R2‬‬
‫(‬
‫= ‪Vc = VR2 = I2 R2 ⇒ I2‬‬
‫‪E = Vc + VR1 = Vc + I1 R1 = Vc + (I2 + Ic ) R1 = Vc +‬‬
‫לפי הגדרה ‪:Ic = cV̇c‬‬
‫(‬
‫)‬
‫‪Vc‬‬
‫‪E = Vc +‬‬
‫‪+ cV̇c R1‬‬
‫‪R2‬‬
‫‪R2‬‬
‫‪R1 R2‬‬
‫‪E‬‬
‫‪− Vc = V̇c‬‬
‫‪c‬‬
‫‪R∫1 + R2‬‬
‫‪R‬‬
‫‪1 + R2‬‬
‫∫‬
‫‪dVc‬‬
‫‪dt‬‬
‫=‬
‫‪−‬‬
‫‪R1 R2‬‬
‫‪2‬‬
‫‪c‬‬
‫‪Vc − E R1R+R‬‬
‫‪R1 +R2‬‬
‫‪2‬‬
‫(‬
‫)‬
‫‪R2‬‬
‫‪t‬‬
‫‪− R1 R2 + Const = ln Vc − E‬‬
‫‪R1 + R2‬‬
‫‪c‬‬
‫‪R1 +R2‬‬
‫(‬
‫)‬
‫‪R R‬‬
‫‪R2‬‬
‫)‪−t/( R 1+R2 c‬‬
‫‪1‬‬
‫‪2‬‬
‫‪Vc (t) = V0 e‬‬
‫‪+E‬‬
‫‪R1 + R2‬‬
‫א‬
‫‪RC 5305‬‬
‫נציב תנאי התחלה שבזמן אפס אין שום מטען לכן אין שום מתח‪:‬‬
‫‪R2‬‬
‫‪R1 + R2‬‬
‫(‬
‫)‬
‫‪R R‬‬
‫)‪−t/( R 1+R2 c‬‬
‫‪1‬‬
‫‪2‬‬
‫‪Vc (t) = E 1 − e‬‬
‫הזרם מוגדר לפי ‪:Ic = cV̇c‬‬
‫‪1‬‬
‫‪R1‬‬
‫‪Ic (t) = Ee−t/R1 c‬‬
‫נציב בזמן אפס‪:‬‬
‫‪E‬‬
‫‪R1‬‬
‫= )‪Ic (0‬‬
‫ב‪ .‬נמצא תנאי התחלה‬
‫‪ER2‬‬
‫‪R1 + R2‬‬
‫= ‪Vc (0) = VR2 = I2 R2‬‬
‫חוק מתחים‪:‬‬
‫‪Vc = −VR2 = −Ic R2 = −cR2 V̇c‬‬
‫‪ER2 −t/R2 c‬‬
‫‪e‬‬
‫= ‪Vc (t) = V0 e−t/R2 c‬‬
‫‪R1 + R2‬‬
‫הזרם מוגדר לפי ‪:Ic = cV̇c‬‬
‫‪E‬‬
‫‪e−t/R2 c‬‬
‫‪R1 + R2‬‬
‫‪Ic (t) = −‬‬
‫נציב בזמן אפס‪:‬‬
‫‪E‬‬
‫‪R1 + R2‬‬
‫ב‬
‫‪Ic (0) = −‬‬
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)