A 100 F capacitor is connected in a series with a10 resistor and a 4.5V battery

Question

A 100 F capacitor is connected in a series with a10 resistor and a 4.5V battery that can be turned off and on.The capacitor is initially fully charged:
A) When will the voltage drop across the 10 resistor beequal to 1.5V after the switch is closed? B) If the battery is switched off after the capacitor is fullycharged and the capacitor then discharged through the circuit, howlong will it take for the voltage across the 10 resistor todrop to 1V ? I was provided with an incorrect answer on my last post, but Ihaven't figured out the correctanswer.        Not a text problem A 100 F capacitor is connected in a series with a10 resistor and a 4.5V battery that can be turned off and on.The capacitor is initially fully charged:
A) When will the voltage drop across the 10 resistor beequal to 1.5V after the switch is closed? B) If the battery is switched off after the capacitor is fullycharged and the capacitor then discharged through the circuit, howlong will it take for the voltage across the 10 resistor todrop to 1V ? I was provided with an incorrect answer on my last post, but Ihaven't figured out the correctanswer.        Not a text problem

Explanation / Answer

Capacitance C = 100 F = 10 ^ -4 F resistance R = 10 ohm Initial voltage V = 4.5 V (A). voltage after time t is V ' = 1.5 V from the relation V ' = V [ 1- e -t / RC] substitue values we get ( 1/ 3 ) =  [ 1- e-t / RC ] solving we get time t value (B).   voltage after time t is V '= 1 V from the relation V ' = V [ 1- e -t' / RC] substitue values we get ( 1/ 4.5 ) =  [1- e -t' / RC ] solving we get time t ' value          from the relation V ' = V [ 1- e -t' / RC] substitue values we get ( 1/ 4.5 ) =  [1- e -t' / RC ] solving we get time t ' value

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