72. A 1.50-F capacitor is connected to a 12.0-V battery for along time, and then

Question

72. A 1.50-F capacitor is connected to a 12.0-V battery for along time, and then is disconnected. The capacitor briefly runs a1.00W toy motor for 2.00sec. After this time, a. by how much has the energy stores in the capacitordecreased? b. what is the voltage across the plates? c. how much charge is stored on the capacitor d. how much longer could the capacitor run the motor, assumingthe motor ran at full power until the end? 72. A 1.50-F capacitor is connected to a 12.0-V battery for along time, and then is disconnected. The capacitor briefly runs a1.00W toy motor for 2.00sec. After this time, a. by how much has the energy stores in the capacitordecreased? b. what is the voltage across the plates? c. how much charge is stored on the capacitor d. how much longer could the capacitor run the motor, assumingthe motor ran at full power until the end?

Explanation / Answer

   Energy stored on acapacitor   U   =   (1/2)* C* V2   =   0.5 *1.50 *12.02   =   108   J    a.   Energy used bymotor   E   =   Power* time   =   1.00 *2.00   =   2.0   W    b.   Remainingenergy   U'   =   U   -   E   =   108-2   =   106   J          U'   =   0.5* C * V'2          106   =   0.5* 1.5 *V'2      =>   V'   =   (212/1.5)   =   11.89   V    c.   Charge   Q   =   C* V   =   1.5 *12   =   18   C    d.   Time the motor could berun   t   =   powerremaining / power   =   106 /1   =   106   s       106   =   0.5* 1.5 *V'2      =>   V'   =   (212/1.5)   =   11.89   V    c.   Charge   Q   =   C* V   =   1.5 *12   =   18   C    d.   Time the motor could berun   t   =   powerremaining / power   =   106 /1   =   106   s

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