Unformatted text preview: so
recall that even though the currents all have labeled directions, these are only
assumptions and a negative answer for the current simply means the assumed direction
was incorrect.
Apply Kirchhoff’s junction rule to the node at B: currents going out are equal to currents
going in:
I3 = I1 + I2 (eq.1) Apply Kirchhoff’s loop rule to the loop ABEF. Recall that a resistor is a voltage drop
when the current is the same direction of the loop.
 R1I1  R3I3  ε2 + ε1 = 0
 6.0Ω⋅I1  3.0Ω⋅I3  2V + 10V = 0
 6.0Ω⋅I1  3.0Ω⋅I3 + 8V = 0 (eq.2) Apply Kirchhoff’s loop rule to the loop CBED Solutions will be posted at www.prep101.com/solutions 25 ©Prep101 Phys1026 April Exam Solutions  R2I2  R3I3  ε2 + ε3 = 0
 6.0Ω⋅I2  3.0Ω⋅I3  2V + 10V = 0
 6.0Ω⋅I2  3.0Ω⋅I3 + 8V = 0 (eq.3) Comparing equations 2 and 3, we can see that I1 and I2 are equal, but we can also
determine this applying Kirchhoff’s loop rule to the loop ACDF:
 R1I1 + R2I2  ε3 + ε1 = 0
 6.0Ω⋅I1 + 6.0Ω⋅I2 – 10V + 10V = 0
 6.0Ω⋅I1 + 6.0Ω⋅I2 = 0
I1 = I2 (eq.4) Substitute (eq.4) into (eq.1)
I3 = I1 + I2
I3 = I1 + I1
I3 = 2I1 (eq.5) Substitute (eq.5) into (eq.2)
6.0Ω⋅I1  3.0Ω⋅I3 + 8V = 0
6.0Ω⋅I1  3.0Ω⋅2I1 + 8V = 0
12.0Ω⋅I1 + 8V = 0
I1 = 2/3A
The rest of the currents fall out…
I2 = 2/3A
I3 = 4/3A
Since all the currents were positive, this means the assumed directions were correct.
Question 2: Solution: ξ 10V
= 0.05 mA
R 200 k Ω
Q = CV = (3⋅106 F) ⋅ (10 V) = 30 μC
τ = RC = (200⋅103 Ω) ⋅ (3⋅106 F) = 0.6 s
I= = Question 3:
Solution: Q1 = C1V and Q2 = C2V
Q1 = 1⋅104 C Q2 = 5⋅105 C
Solutions will be posted at www.prep101.com/solutions 26 ©Prep101 Phys1026 April Exam Solutions The effective resistance is:
1
1
1
=
+
⎯
⎯→ R eq = 50 Ω
R eq 100 Ω 100 Ω
Δ V 10 V
I=
=
= 0.2 A
R
50 Ω Use equivalent resistance Req = 50 Ω and capacitance Ceq = C1 + C2 = 15.0 μF
τ = RC = 7.5⋅104 s
Question 4:
Solution: ε1 = ε2 = 10 V, ε3 = 2.0 V
R1 = R2 = 6.0 Ω, R3 = 3.0 Ω
Node at B: I3 = I1 + I2 (eq.1)
ABEF: 0 =  R1I1  R3I3  ε2 + ε1 = 6I1 + 3I3 (eq.2)
ACDF: 0 =  R1I1 + R2I2  ε3 + ε1 = 6(I2 – I1)+8 (eq.3)
From (2): I3 = 2I1
Substitute into (1): I1 + I2 = 2I1 ⇔ I2 = 3I1
Substitute into (3): 6(3I1 – I1)+8 = 0 ⇔ I1 = 1/3 A ⇒ I2 = 1 A I3 = 2/3 A
Question 5: Solution:
For A:
After a long time the capacitor branch will carry negligible current. The current flow is
shown in the figure (a) below. To find the voltage at point a, we use Kirchhoff’s voltage rule: Solutions will be posted at www.prep101.com/solutions 27 ©Prep101 Phys1026 April Exam Solutions Similarly at point b: Therefore, the voltage across the fully charge capacitor is: For B:
We suppose the battery is pulled out leaving an open circuit. We are left with figure (b),
which can be reduced to equivalent circuits (c) and (d). From (d), we can see that the capacitor sees 3.6 Ω during its discharge.
and the fact that V = CQ or According to, Subst...
View
Full
Document
This document was uploaded on 03/21/2014 for the course PHYSICS 1402 at UWO.
 Winter '14
 Prof.PaulineBarmby
 Physics

Click to edit the document details