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Prep101comsolutions 26 prep101 phys1026 april exam

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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⋅10-6 F) ⋅ (10 V) = 30 μC τ = RC = (200⋅103 Ω) ⋅ (3⋅10-6 F) = 0.6 s I= = Question 3: Solution: Q1 = C1V and Q2 = C2V Q1 = 1⋅10-4 C Q2 = 5⋅10-5 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⋅10-4 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...
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This document was uploaded on 03/21/2014 for the course PHYSICS 1402 at UWO.

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