For the capacitor network shown in the figure below , the potential difference a
ID: 2048298 • Letter: F
Question
For the capacitor network shown in the figure below , the potential difference across ab is 12.0 { m V}.http://session.masteringphysics.com/problemAsset/1263956/1/1019818.jpg
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Find the total energy stored in this network.
U = ______
Find the energy stored in the 4.80 {nF} capacitor.
U_{4.80nF} = _________________
Explanation / Answer
The capacitors 8.6 µF & 4.8µF are in series. Their equivalent capacitance C (say) So, 1/C = 1/8.6 + 1/4.8 => C = 8.6*4.8/(8.6 + 4.8) = 3.0806 µF Again capacitors 6.2µF & 11.8µF are in series. Their equivalent capacitance is C'(say) So, 1/C' = 1/6.2 + 1/11.8 C' = 6.2*11.8/(6.2 + 11.8) = 4.0644 µF Now C' & 3.5µF are in parallel. The equivalent of these two is C'' = C' + 3.5 = 4.0644 + 3.5 =7.5644µF Now C & C'' are in series Thus equivalent capacitance of the circuit isCtotal(say) 1/Ctotal = 1/C + 1/C'' Ctotal = C*C''/(C+C'') = 3.0806*7.5644/(3.0806 +7.5644) = 2.1891µF So, total potentialenergy stored in network = 1/2 * Ctotal*V2 =(1/2)*2.1891*122 = 157.6152µJ Now the charge stored in the network is q = CtotalV= 2.1891*12 = 26.2692µC this will be the charge on each of capacitors 8.6µF &4.8µF [as they are in series] [this charge is devided in parallal branches] So, the charge on capacitor 4.8µF is 26.2692µC So the energy stored in4.8µF is (1/2)q2/C = (1/2)*26.26922/4.8 =71.8824µJ [Note: p.e in a capacitor is 1/2 CV2=1/2q2/C]