Clayton College & State University

 

February 3, 2003

 

Physics 1112 – Quiz 3a

 

 

Name  ____SOLUTION_________________________________

 

 

  1. Consider the network of capacitors shown below. All the capacitors are identical and have the capacitance C = 5.00 mF.

 

    1. Calculate the equivalent capacitance of the network.

 

C345 = C3 + C4 + C5 = 5.00 mF +  5.00 mF + 5.00 mF= 15.00 mF

 

1/C12345 = 1/C1 + 1/C2 + 1/C345= 1/(5.00 mF) +  1/(5.00 mF) + 1/(15.00 mF) = 7/(15.00 mF)

 

C12345 =(15/7) mF = 2.14 mF

 

    1.  Given that the battery provides 12.0 V of potential difference, find how much work was done by the battery in order to charge the network.

 

ES = ½ C (DV)2 = ½ (2.14mF) (12.0 V)2 = 154 mJ