Clayton College & State University

 

February 3, 2003

 

Physics 1112 – Quiz 3b

 

 

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.

 

1/C12 = 1/C1 + 1/C2 = 1/(5.00 mF) +  1/(5.00 mF) = 2/(5.00 mF)

 

C12 = 5/2 mF

 

C1234 = C12 + C3 + C4= 5/2 mF +  5.00 mF + 5.00 mF = 25/2 mF  = 12.5 mF

 

1/C12345 = 1/C1234 + 1/C5 = 2/(25 mF) + 1/(5.00 mF) = 7/(25 mF)

 

C12345 = 25/7 mF = 3.57 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 (3.57 mF) (12.0 V)2 = 257 mJ