Capacitor problem… in series and parallel?

Question

Using this figure of a circuit:http://session.masteringphysics.com/problemAsset/1...A) What is the charge on each capacitor in the figure ?Enter your answers numerically separated by commas.B) What is the potential difference across each capacitor in the figure?Enter your answers numerically separated by commas.I'm so lost 🙁 please help

Johathan Wehner · Accepted Answer

V2 = V*C3/(C2+C3 ) = 9*6/10 = 5.4V ....Q2 = V2*C2 = 5.4*4 = 21.60 μCV3 = V*C2/(C2+C3) = 4*9/10 = 3.6V ....Q3 = V3*C3 = 3.6*6 = 21.60 μCV1 = V = 9.0 V....................Q1 = V*C1 = 5*9.0 = 45 μC

Johathan Wehner · Answer

Key ideas are charge-voltage relation Q = C.VC1 is 5 uF with 9V p.d. across it hence Q1 = 5x10^-6 X 9 = 45x10^-6 CoulombsThe series arrangement is a bit more challenging and we can use a number of differentstarting points. Lets begin with the idea that the series capacitors each have the SAME charge i.e Q2 = Q3 so the voltages V split in inverse relation to their capacitances.The total capacitance 1/C = 1/C2 + 1/C3 =2.4 uF giving Q2 =Q3 = 2.4*9x10^-6 = 21.6x10^-6 Coulombsand finally V2 = 21.6/4 = 5.4 V and V3 = 21.6/6 = 3.6 V.(verify that 5.4 + 3.6 =9 as required and 5.4/3.6 = 6/4 as noted above!!)

Johathan Wehner · Answer

The two in series are equivalent to 6•4/10 = 2.4 µFC1 has 9v on it, and Q1 = CV = 9•5 = 45 µC chargeThe series combo has 9v across the pair, and the charge is 9•2.4 = 21.6 µC, and that charge is on both of the pair. V2 = Q/C = 21.8 / 4 = 5.4 voltsV3 = 21.8 / 6 = 3.6 volts

Johathan Wehner · Answer

connect 2 x a million micro farad capacitors in sequence this may will equivalent a .5 farad capacitor. connect those in Parallel with a a million microfarad capacitor and voile! you've a a million.5 microfarad cap

Haylie Ernser DVM

Capacitor problem… in series and parallel?

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