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# A battery that has an emf and a resistor with a

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Unformatted text preview: ow we analyze the simple circuit with a resistor, R, and inductor, L, in series with a battery.   Kirchhoff ’s loop rule VBattery + VR + VL = 0 dI VBattery − IR − L = 0 dt Physics 2B - Summer Session 2 - C. Palmer August 28, 2013 22   Now we analyze the simple circuit with a resistor, R, and inductor, L, in series with a battery.   Kirchhoff ’s loop rule   Plug in assumed solution dI VBattery − IR − L = 0 dt ( I (t ) = I 0 1 − e − tτ ) Physics 2B - Summer Session 2 - C. Palmer August 28, 2013 23   Now we analyze the simple circuit with a resistor, R, and inductor, L, in series with a battery.   Exact constants dI VBattery − IR − L = 0 dt ( I (t ) = I 0 1 − e I0 = − tτ ) VBattery L ; τ= R R Physics 2B - Summer Session 2 - C. Palmer August 28, 2013 24 A solenoid having an inductance of 6.30µH is connected in series with a 1.20kΩ resistor. If a 14.0V battery is connected across the pair, how long will it take for the current through the resistor to reach 80.0% of its final value? A)  310 Ms B)  8.4 s C)  0.31 s D)  8.4 ns E)  1.2 ns ( I (t ) = I 0 1 − e − tτ ) VBattery L I0 = ; τ= R R Physics 2B - Summer Session 2 - C. Palmer August 28, 2013 25 A solenoid having an inductance of 6.30µH is connected in series with a 1.20kΩ resistor. If a 14.0V battery is connected across the pair, how long will it take for the current through the resistor to reach 80.0% of its final value? A)  310 Ms B)  8.4 s C)  0.31 s D)  8.4 ns ( 0.8 * I 0 = I 0 1 − e 0.2 = e − tτ ; E)  1.2 ns Physics 2B - Summer Session 2 - C. Palmer August 28, 2013 − tτ ) L τ= R 26   Energy delivered to an inductor is similar to power dissipated in a resistor… except it is stored and not radiated. dU = I (t ) dε L UStored = ∫ I0 0 12 I ( LdI ) = LI 0 2 Physics 2B - Summer Session 2 - C. Palmer August 28, 2013 27   Start from the energy density in an inductor.   Then recall the inductance in a solenoid. 12 U L = LI 0 2 2 UStored LI 0 uL = = V 2V L = µ0 n V Physics 2B - Summer Session 2 - C. Palmer 2 August 28, 2013 28   Start from the energy density in an inductor.   Then recall the inductance in a solenoid. 2 0 U L LI uL = = V 2V uSolenoid L = µ0 n V 2 LI µ0 n VI 1 22 = = = µ0 n I 0 2V 2V 2 2 0 2 Physics 2B - Summer Session 2 - C. Palmer 2 0 August 28, 2013 29   Start from the energy density in an inductor.   Then recall the inductance in a solenoid.   Put the energy density in terms of the magnetic field uSolenoid B = µ0 nI 0 2 Solenoid 1 B 22 = µ0 n I 0 = 2 2 µ0 B uB = 2 µ0   This is actually a general statement   There is energy in the field Physics 2B - Summer Session 2 - C. Palmer 2 August 28, 2013 30 L   Let us exam a charged capacitor, C, with an...
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