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Transcript
```3ʤʬʠʹ
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
=
=
I =
= 0.485 A
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)
4ʤʬʠʹ
```
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