A battery that has an emf and a resistor with a

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online