Physics 1E03 – Practice Test 2 Solutions
加拿大物理Final exam代考 In the circuit shown below, the resistances are R1 = 4.0 Ω and R2 = 4.0 Ω, and the ideal batteries have emfs e1 = 2.0 V
Part B (Problems): 加拿大物理Final exam代考
1.In the circuit shown below, the resistances are R1 = 4.0 Ω and R2 = 4.0 Ω, and the ideal batteries have emfs e1 = 2.0 V and e2 = e3 = 6.0 V. Apply Kirchhoff’s circuit rules (junction and looprules):
Note: Include the correct sign to indicate direction relative to the arrows shown.
A.Write junction and loops rule Junction: I1 +I 2 – I3 =0
I3 = I1 + I 2 (1)
Left Loop:
e1 – I3 R2 – e 2 – I1 R1 = 0
e1 – ( I1 + I 2 ) R2 – e 2 – I1 R1 = 0(2)
Right Loop:
e 2 + I3 R2 + I2 R1 – e 3 = 0
e 2 + (I1 + I2 ) R2 + I2 R1 + e3 = 0(3)
B.Find the magnitude and direction of I1, I2 and I3.
(2)Þ 2 – (I1 + I 2 ) (4)– I1 (4) – 6 = 0
1 + 2I1 + I 2 = 0(4)
(3) Þ (I1 + I 2 ) (4) + 2 I 2 (2) = 0
I1 + 2I 2 = 0 I1 = – 2I 2 (5)
1 + 2( –2I 2 ) + I 2 = 0
I 2 =1 3 = 0.33 A
I1 =– 2 (1 /3) =-2/3 I1 = + 2 3 = + 0.67 A 加拿大物理Final exam代考
I3 = – 2/ 3 + 1 /3= – 1 /3 I3 = + 1 3
C.What is the potential difference Va –Vb?
Va – I3 R2 – e 2 = Vb2
Va – Vb = I3 R2 + ε 2
Va – Vb = (-1 /3)( 4) + 6 = 14/3 V `= 4.67 V
Va – Vb = 14 /3 V = 4.67 V
2.In the RC circuit of the diagram shown, ε = 9.00 V, C = 10.0 μF, R1 = 3.00 Ω, 加拿大物理Final exam代考
R2 = 15.0 Ω and R3 = 12.0 Ω, the switch has been closed for a very long time so that the capacitor is fully charged.
A.Find the time constant when the switch isopen 
τ = C Reff = C (R1 + R2 )
τ = 10.0×10–6 F (3.0 +15.0)Ω = 1.8×10– 4 S
B.Find the charge qO on the capacitor the instant the switch is opend.
At t =0, qo is the same as q¥ when the switch was closed and the capacitor was fully charged. Fully charged capacitor acts like an open switch, so:
C.Write an expression for the charge on the capacitor as a function of time t >0.
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