Sarahisme Posted May 25, 2005 Author Posted May 25, 2005 this one i am not so sure about...or not confident enough to try and explain it
Douglas Posted May 25, 2005 Posted May 25, 2005 The battery voltage is constant the instant the switch is closed the voltage on the capacitor is 0v thus, the voltage drop across the resistor is equal to the battery voltage since the cap blocks DC, one plate begins to charge. When the charge reaches the battery voltage the voltage drop across the resistor is 0v
mezarashi Posted May 25, 2005 Posted May 25, 2005 I agree with (e) too Here are some mathematical aids. See for yourself which one is true Using the formal definitions introduced in electric circuits: (a) i = C dV/dt Cdv = int (i dt) CV = Q where C = capacitance, V = voltage across capacitor = Vcapacitor, Q = charge on capacitor. Note that Vcapacitor is always less than Vbattery, and so Q/C is always less than Vbattery. i = V/R (Ohm's Law) i = (Vbattery - Vcapacitor)/R i = (Vbattery - Q/C)/R (b) Voltage drop across resistor: delta V = Vbattery - Vcapacitator delta V = Vbattery - Q/C © i = (Vbattery - Q/C)/R (d) Q = charge on capacitor = charge across capacitor = Qcapacitor (e) delta V = Vbattery - Q/C
Sarahisme Posted May 26, 2005 Author Posted May 26, 2005 wait a second there isnt more than one of those options that are correct are there? ie. e is the ONLY correct answer...?? right? cause i think it is...
mezarashi Posted May 26, 2005 Posted May 26, 2005 Yes ^^ only e is true. Follow the arguments. See if the changing the variables in the equations for a, b, c, and d will make the assumptions in the questions hold true.
Sarahisme Posted May 26, 2005 Author Posted May 26, 2005 yeah i thought so, just checking thankss mezarashi
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