Physics 21Fall, 2004Solution to HW-926P44Two 6.0μF capacitors, two 2.2 kΩ resistors, anda 12.0 V source are connected in series. Starting from theuncharged state, how long does it take for the current todrop from its initial value to 1.50 mA?2200Ω2200Ω6.0µF 6.0µF12 VReplace theC’s by equivalent value1Cef=16.0μF+16.0μF⇒Cef=3.0μF,and theR’s by equivalent valueRef=2.2kΩ+2.2 kΩ = 4400 Ω.The time constant isRefCef=(3.0×10-6F)(4400 Ω) = 13.2msAt the instant the current starts, the capacitors are un-charged, so the voltage drop is entirely across the resistors.Hence the initial value of the current isI0=12 V4400 Ω.727 mA.We must solveexp(−t/13.2ms)=1.50 mA2.727 mA=0.55.The result ist=13.×lnµ1.55¶=7.9ms26P80Determine the current in each resistor of the cir-cuit shown in Fig. 26-57.123I4I5I6I12nodeEach current is labelled by the resistor it ﬂows through.node:I5+I4=I6+I12loop 1:5 + 10−5I5−
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