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ricotico8 ricotico8
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9 years ago
The capacitive network shown is assembled with initially uncharged capacitors. A potential difference,  is applied across the network. The switch S in the network is kept open throughout. In Figure 18.12, the total energy stored in the seven capacitors, in mJ, is closest to:
      
144
      
96
      
72
      
120
      
48
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#1
Educator
9 years ago
the 4μF, 8μF and 12μF are in parallel so the equivalent capacitance is

C1 = 4μF + 8μF + 12μF

C1 = 24μF

now C1 and the 6μF are in series so the equivalent capacitance C2 is

1/C2 = 1/24μF + 1/6μF

C2 = 4.8 μF

the 9μF and 15μF are in parallel so the equivalent capacitance C3 is

C3 = 9μF + 15μF

C3 = 24μF

now

Vab = Vac + Vcb

100 V = q1/C3 + q1/16μF

100 V = q1(1/24μF + 1/16μF)

q1 = 960 μC

Vab = Vad + Vdb

100 V = q2/6μF + q2/C1

100 V = q2(1/6μF + 1/24μF)

q2 = 480 μC

and

Vbc = Vcd + Vdb

Vcd = Vbc - Vdb

Vcd = 960C/16μF - 480μC/24μF

Vcd = 40 V
Ravinod
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