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PHY2054 - Fall 2006 Exam 2 Key

# PHY2054 - Fall 2006 Exam 2 Key - Question 1 R1 U2 R3 U1 R4...

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Question 1: R1 R2 R3 R4 C1 U3 U1 U2 The current circuit of voltage supplies, resistors, and capacitors is in steady state. The resistors have the following values: R 1 = 15 , R 2 = 25 , R 3 = 14 , R 4 = 26 . The capacitor has a capacitance of C 1 = 5 μF and the batteries have the following voltages: U 1 = 10 V , U 2 = 20 V , U 3 = 35 V . Calculate the magnitude of the voltage (in V) across C 1. Answer 1: We realize that the capacitor stops any current flow through R 3 and U 3 . That means that the potential difference across R 3 is zero. The only closed loop with current is the outer loop. The potential differences around the loop starting at a and going clockwise have to add up to 0: Σ∆ V = R 1 I + U 2 + R 2 I + R 4 I U 1 = 0 I = U 1 U 2 R 1 + R 2 + R 4 = 0 . 1515A The voltage diffences around the right loop also have to add up to 0: V C 1 U 3 + U 2 + IR 2 = 0 V C 1 = U 3 U 2 R 2 I = 18 . 788V This can be checked for consistency using the left loop: V C 1 U 3 IR 1 + U 1 IR 4 = 0 V C 1 = U 3 U 1 +( R 1 + R 4 ) I = 18 . 788V 1

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Question 2 R1 R2 R3 R4 C1 U1 The circuit shown in the left figure has the following components: U 1 = 6V, R 1 = 10 , R 2 = 15 , R 3 = 100 , R 4 = 25 , and C 1 = 1mF = 10 3 F. Assume the switch is closed and the capacitor is fully charged. What is the voltage drop across R 2 ? Answer 2: In steady state the circuit is a simple series of resistors R 1 , R 2 , and R 4 . The capacitor does not allow any current to flow through R 3 in steady state. The voltage drop is: V R 2 = R 2 U 1 R 1 + R 2 + R 4 = 1 . 8V Question 3 Look again at the figure in Question 2. If we open the switch after the capacitor was fully charged, how long does it take for the charge on C 1 to drop to 25% of its initial value? (For this problem it is not necessary to have solved Q2) Answer 3: When we open the switch the charge on the capacitor can only flow through the resis- tors R 3 , R 2 , and R 4 . The time constant is then: τ = C 1 × ( R 3 + R 2 + R 4 ) = 0 . 14s To calculate the time we need the following equation: 1 4 = e t / τ t = τ ln ( 0 . 25 ) = 0 . 194s 2
Question 4 A charged particle with mass m = 1 . 67 · 10 27 kg and charge q = 1 . 6 · 10 19 C, initially at rest, will be accelerated from west to east over a distance of 0 . 5m by an electric field of 3kV / m. It then enters a region with a magnetic field of 0 . 354T pointing from north west to south east. What is the cyclotron radius of the motion of the particle in the field? Answer 4: The kinetic energy after being accelerated by the electric field is equal to the potential energy the particle had before it was accelerated: E kin = 1 2 mv 2 = qV = E pot v = radicalbigg 2 qV m = 536km / s The magnetic force on the particle is given by: F mag = qvB sin45 = qvB 2

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PHY2054 - Fall 2006 Exam 2 Key - Question 1 R1 U2 R3 U1 R4...

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