HW11solutions - Physics 213 30.19. Homework 11 Solutions...

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Physics 213 Homework 11 Solutions Spring 2008 30.19. IDENTIFY: Apply Kirchhoff’s loop rule to the circuit, sketched below. i ( t ) is given by Eq.(30.14). SET UP: di dt is positive as the current increases from its initial value of zero. EXECUTE: ± ² v R ² v L = 0 , or ² iR ² L di dt = 0, so i = R 1 ² e ² ( R / L ) t () (a) Initially ( t = 0), i = 0 so ² L di dt = 0 , and di dt = L = 6.00 V 2.50 H = 2.40 A/s (b) ² iR ² L di dt = 0 (Use this equation rather than Eq.(30.15) since i rather than t is given.) Thus di dt = ² iR L = 6.00 V ² (0.500 A)(8.00 ³ ) 2.50 H = 0.800 A/s (c) i = R 1 ² e ² ( R / L ) t = 6.00 V 8.00 ³ ´ µ · ¸ ¹ 1 ² e ² (8.00 ³ /2.50 H)(0.250 s) = 0.750 A(1 ² e ² 0.800 ) = 0.413 A (d) Final steady state means t ±² and di dt ± 0, so ³ ´ iR = 0. , and thus i = R = 6.00 V 8.00 ² = 0.750 A EVALUATE: Our results agree with Fig.30.12 in the textbook. The current is initially zero and increases to its final value of / R . The slope of the current in the figure, which is di/dt , decreases with t . 30.61. IDENTIFY and SET UP: The current grows in the circuit as given by Eq.(30.14). In an R - L circuit the full emf initially is across the inductance and after a long time is totally across the resistance. A solenoid in a circuit is represented as a resistance in series with an inductance. Apply the loop rule to the circuit; the voltage across a resistance is V = IR . EXECUTE: (a) In the R - L circuit the voltage across the resistor starts at zero and increases to the battery voltage. The voltage across the solenoid (inductor) starts at the battery voltage and decreases to zero. In the graph, the voltage drops, so the oscilloscope is across the solenoid. (b) At t the current in the circuit approaches its final, constant value. The voltage doesn’t go to zero because the solenoid has some resistance R L . The final voltage across the solenoid is IR L , where I is the final current in the circuit. (c) The emf of the battery is the initial voltage across the inductor, 50 V. Just after the switch is closed, the current is zero and there is no voltage drop across any of the resistance in the circuit. (d) As t , ´ IR ´ IR L = 0 ± =50V, and from the graph IR L =15V (the final voltage across the inductor), so IR = 35 V and I = (35 V)/ R = 3.5 A (e) IR L = 15 V, so R L = (15 V)/(3.5 A) = 4.3 ± ² V L ² iR = 0, where V L includes the voltage across the resistance of the solenoid.
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HW11solutions - Physics 213 30.19. Homework 11 Solutions...

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