Charge distribution method
Star/ delta network
Wheatstone bridge cases
If C1/C2 = C3/C4 the eliminate C5.
If in a Wheatstone bridge each capacitor is C then C eq = C.
Charge distribution cases
Apply charge distribution of the circuit is symmetrical. In symmetrical circuits charge entering a branch = charge leaving an identical branch.
If two capacitors C1 and C2 charged to V1 and V2 are joined together then common potential is
Vcommon = V1 C1 + V2 C2/C1 + C2 = Q1 + Q2/C1 + C2
Charge on capacitors after joining Q1/Q2 = C1/C2
Q1 = (Q1 + Q2) C1/(C1 + C2 Q2 = (Q1 + Q2) C2/C1 + C2
Loss in energy when tow capacitors C1 and C2 charge to V1 and V2 are joined together is
?E = C1C2/2 (C1 + C2) (V1 - V2)2
If a dielectric slab in a capacitor is being introduced in the rigidly held plates then the force required to insert the slab is
F = 1/2 V2 dC/dx
Growth transient or charging of a capacitor:- if Q is charge at any instant across the capacitor then
Q = Q0 (1 - e - t/ Rc) where Q = CV0
VR = V0 e -t/RC I = dQ/dt = Ra/RC e - I/Rc
Time constant τ is that during which a capacitor charges to 63% of its maximum value of charge.
Discharging of a capacitor (decay transient)
Q = Q0 e -t/RC' VR = V0 (1 - e -t/Rc)
Time constant τ = RC is defined as the time during which capacitor discharges to 36% of maximum charge.
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