What is the potential difference across each capacitor in the figure?
http://session.masteringphysics.com/problemAsset/1071530/4/30.P60.jpg What is the charge on each capacitor in the figure ? Q_1, Q_2, Q_3 = ? What is the potential difference across each capacitor in the figure? v1, v2, v3 = ? ty in advance :)
What will be the potential difference across the capacitor in an R-C series circuit?
After 5 RC time constant the potential across Capacitor is full DC Voltage connected
How are the capacitors connected through series/parallel? What is the potential difference across each capacitor?
KVL: +E1 - Vc1 + E2 - Vc2 = 0Vc1 + Vc2 = E1 + E2Total charge flown is same => Qc1 = Qc2 => C1*Vc1 = C2*Vc2 => Vc1 = C2*Vc2/C1C2*Vc2/C1 + Vc2 = E1+E2Vc2 = (E1+E2)*C1/(C1+C2)Vc1 = (E1+E2)*C2/(C1+C2)
Find (a) total capacitance (b) potential difference across each capacitor and...?
a) Total capacitance = 5.0F (capacitors in parallel add) b) Potential difference across each capacitor = 100V (parallel capacitors have the same voltage) c) Charge stored in each capacitor; Q = CV where Q = charge, C = capacitance, and V = voltage across the capacitor. For the 2.0F capacitor, Q = CV = 2 x 100 = 200 coulombs. For the 3.0F capacitor, Q = CV = 3 x 100 = 300 coulombs.
Find the resulting potential difference across each capacitor after they are connected?
While connecting to the battery, both capacitors are equally charged (serial connection), assumed the charge is Q. Q/C1 +Q/C2 = U, we have Q = U / (1/C1+1/C2) = 18/(1/4+1/12) = 54 μCoulomb After they are disconnected to the battery and reconnected to each other, the potential difference across both of them are the same (parallel connection), assume the potential is Un. And we have charge conservation: C1 Un + C2 Un = Q + Q Un = 2 Q / (C1+C2) = 2 *54/(4+12) = 6.75 V initial energy stored in capacitors Ei = 0.5 (Q^2/C1+ Q^2/C2) = 0.5 Q^2 (1/C1+ 1/C2) = 0.5*54^2*(1/4+1/12) = 486 μJ final energy stored in capacitors Ef = 0.5 (C1 U^2 + C2 U^2) = 0.5 U^2 (C1+ C2) = 0.5*6.75^2*(4+12) = 364.5 μJ in the first case we used E=0.5 Q^2/C, in the second E=0.5 C U^2, because we know the Q in the first case, and U in the second. The choices make the calculation convenient.
Physics: Calculate the potential difference across each capacitor.?
Because they are in series, each capacitor has the same charge Q. We know that generally Q = CV so Q/C = V Therefore: Q/C1 + Q/C2 = 10V => QC1+ QC2 = 10C1C2 = 7e-12 => Q = 7e-12(C1+C2) = 3.68e-6 C A) Calculate the potential difference across each capacitor. Vc1 = Q/C1 = 3.68e-6/0.5e-6 = 7.37V <----- Vc2 = Q/C2 = 3.68e-6/1.4e-6 = 2.63V <----- B) Calculate the charge on each capacitor. Q = 7e-12(C1+C2) = 3.68e-6 C <----- C) Calculate the potential difference across each capacitor assuming the two capacitors are in parallel. Since they are in parallel, they have the same voltage V and the charge Q is distributed accordingly C1V+C2V = Q1+Q2 = Q => V = Q/(C1+C2) = 3.68E-6/[(1.4+0.5)*1e-6] = 1.94v <----- D) Calculate the charge on each capasitor assuming the two capacitors are in parallel. Q1 = C1*V = 0.5e-6*1.94V = 0.970e-6 C <----- Q2 = C2*V = 1.4e-6*1.94V = 2.71e-6 C <----- Notice that Q1+Q2 = 0.97e-6+2.71e-6 = 3.68e-6 C
Why is potential same across the ends for capacitor connected parallel to each other? Explain theoretically?
Each capacitor in parallel combination has equal potential difference between its two plates. Justify the statement.In parallel combination all the capacitors get same potential difference (V) because all the capacitors are connected across the same terminals of battery.In Parallel combination, the left plate of each capacitor is connected to the positive terminal of the battery by a conducting wire. In the same way, the right plate of each capacitor is connected to the negative terminal of the batteryTherefore, each capacitor connected to a battery of voltage V has the same potential difference V across it. i.e.,V1= V2= V3= VFor moreEach capacitor in parallel combination has equal potential difference between its two plates. Justify the statement.
Why is voltage different across capacitors in series?
When capacitors are connected in series, each must hold same amount of charge. This result comes from Kirchoff’s current law since being in series they are charged/ discharged by same amount of current. Now, imagine two capacitors C1 & C2 connected in series across a battery V. Let the individual voltages be V1 and V2 respectively.Since charges across both capacitors have to be equal -> Q = C1V1 = C2V2.Also by Kirchoff’s voltage law, V=V1+V2.Solving, V= Q[C1C2/(C1+C2)].This term C1C2/(C1+C2) is called equivalent capacitance of the circuit (C). As you can see by substituting any value for C1 and C2 , C will be lower than smallest capacitance.
What will be the potential difference across 2 microfarad when two capacitors C1=2 microfarad and C2=4 microfarad are connected in series and a potential difference of 1200V is applied across it?
[math]C1 = 2\mu\,{F}[/math][math]C2 = 4\mu\,{F} [/math][math]Ceq = \dfrac{2\times 4}{2 + 4} =\dfrac{4}{3}\mu\,{F}[/math][math]\text{total charge Qt} = \dfrac{4}{3}\times 1200 = 1600\mu\,{Coulombs}[/math][math]\text{ for series circuit Qt = Q1 = Q2}[/math][math]\text{Voltage across C1} = \dfrac{Qt}{C1}= \dfrac{1600}{2}= 800 V[/math]