12, -8 points YF14 26.P.041. My Notes Ask Your A 4.55HF capaator that is initial
ID: 778581 • Letter: 1
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12, -8 points YF14 26.P.041. My Notes Ask Your A 4.55HF capaator that is initially uncharged is connected in series with a 7.70-k12 resistor and an emf source with -105 v and negligible internal resistance. (a) Just after the circuit is campleted, what is the voltage drop across the capacitor? (b) Just after the circuit is completed, what is the valtage drop across the resistor? (c) Just after the circuit is completed, what is the charge on the capacitor? (d) Just after the circuit is completed, what is the current through the resistor? (e) A long time after the circuit is completed (after many time constants) what are the values of the quantities in parts(aj-(d)? VR- Subrmit Answer Save Progress Practice Another VersionExplanation / Answer
Capacitance = C = 4.55 uF
Resistance = R = 7700 ohm
emf of source = E=105 V negligible internal resistance.
A)Capacitor acts like a straight wire, thus just after the circuit is completed, the voltage drop across the capacitor is zero.
B) Since no voltage drop occurs across the capacitor, all the voltage drops across the resistor. Thus, just after the circuit is completed, the voltage drop across the resistor is105 V.
C) Just after the circuit is completed, the charge on the capacitor is zero.
D)Just after the circuit is completed, the current through the resistor is 105/7700=13.63 mA
E) A long time after the circuit is completed the valuesof the quantities in parts (a)-(d) are
a)After a long time, capacitor acts like a very large resistor (or break). So voltage across the capacitor becomes 105 V .
b) Since all the voltage drops across the capacitor after a very long time thus voltage drop across the resistor is zero
c)The charge across capacitor becomes Q=CV=4.55*10^{-6}*105=477.75 uC
d) After long time,the current through the resistor is zero.
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