How Do You Calculate Voltage Across Vcd in a Capacitor Circuit?

  • Thread starter Thread starter chronicals
  • Start date Start date
  • Tags Tags
    Capacitor
AI Thread Summary
To calculate the voltage across Vcd in a capacitor circuit, first determine the equivalent capacitance (Ceq) and total charge (Qtotal) using the formula Q=CV. In the initial state with the switch open, the equivalent capacitance is found to be 4.00 μF, resulting in a total charge of 840 μC. The potential drops across the capacitors are calculated, leading to Vcd being 70 V after determining the voltage drops from points a to d and a to c. When the switch is closed, the configuration changes, and the charges on the capacitors are recalculated, showing a net charge flow of 315 μC through the switch. Understanding these calculations is crucial for analyzing capacitor circuits effectively.
chronicals
Messages
35
Reaction score
0

Homework Statement


http://img4.imageshack.us/img4/3144/capaeo.jpg




Homework Equations


Q=CV



The Attempt at a Solution


a) i found Ceq=4.00 μF
Qtotal =CV=(4.00 μF) (210 V)=8.40×10−4C. How i can find Vcd=?
b)i have solved this.
c)i have no idea about this.
 
Last edited by a moderator:
Physics news on Phys.org
(a) The switch S is open. The top two capacitors are in series: Ctop = 1/(1/3+1/6) = 2 μF, and the
bottom two are also in series: Cbot = 1/(1/6 + 1/3) = 2 μF. The top and bottom capacitors are in
parallel: Ceq = Ctop + Cbot = 4 μF. The total charge on the capacitors is: Qtot = CeqVab = 840 μC.
Recall that capacitors in parallel split charge. From symmetry, the top capacitors have half the
charge and the bottom capacitors have half the charge. Capacitors in series have the same charge.
Thus, both top capacitors each have 420 μC and both bottom capacitors each have 420 μC. That
is, all four capacitors have exactly the same charge Q = 420 μC. The potential drop from a to d is
thus: Vad = Q/C = (420 μC)/(3 μF) = 140 V, and the potential drop from a to c is: Vac = Q/C =
(420 μC)/(6 μF) = 70 V. Thus, the potential difference between c and d is: Vcd = Vad −Vac = 70 V.



(b) The switch S is now closed. Now the two left capacitors are in parallel: Cleft = 3 + 6 = 9 μF,
and the two right capacitors are in parallel: Cright = 6 + 3 = 9 μF. The left and right sets are in
series: Ceq = 1/(1/9 + 1/9) = 4.5 μF. Recall that for capacitors in series, the potential drops add.
From symmetry, the potential drop across the left capacitors is the same as the potential drop across
the right capacitors: Vleft = Vright = Vab/2 = 105 V. The two right capacitors must have the same
potential drop because they are parallel, and the same goes for the two left capacitors. That is, the
potential drop across each and every capacitor in this configuration is: V = Vad = Vdb = Vac =
Vcb = 105 V.



(c) After the switch is closed, the charge on the top left capacitor is: QTL = CTLVad = 315 μC;
the charge on the top right capacitor is: QTR = CTRVdb = 630 μC; the charge on the bottom
left capacitor is: QBL = CBLVac = 630 μC; and the charge on the bottom right capacitor is:
QBR = CBRVcb = 315 μC. The right side of the top left plate is the negative side, and has charge
−315 μC, and the left side of the top right plate is the positive side, and has charge +630 μC. Thus,
the total charge on the two plates closest to point d is +630 − 315 = 315 μC. Before the switch
closed, there was zero net charge on the two plates closest to d: 420− 420 = 0 μC. So 315 μC must
have flowed through the switch from bottom to top.




I hope this would help you . c ya
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top