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# Capacitance Networks Assignment Help

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Electromagnetism - Capacitance Networks

**Capacitance Networks**

Series/parallel method

Charge distribution method

Star/ delta network

Wheatstone bridge cases

If **C**_{1}/C_{2} = C_{3}/C_{4} the eliminate C_{5}.

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 **C**_{1} and **C**_{2} charged to **V**_{1} and **V**_{2} are joined together then common potential is

**V**_{common} = V_{1} C_{1} + V_{2} C_{2}/C_{1} + C_{2} = Q_{1} + Q_{2}/C_{1} + C_{2}

Charge on capacitors after joining **Q**_{1}/Q_{2} = C_{1}/C_{2}

**Q**_{1} = (Q_{1} + Q_{2}) C_{1}/(C_{1} + C_{2} Q_{2} = (Q_{1} + Q_{2}) C_{2}/C_{1} + C_{2}

Loss in energy when tow capacitors **C**_{1} and **C**_{2} charge to **V**_{1} and **V**_{2} are joined together is

**?E = C**_{1}C_{2}/2 (C_{1} + C_{2}) (V_{1} - V_{2})^{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 V**^{2} dC/dx

Growth transient or charging of a capacitor:- if Q is charge at any instant across the capacitor then

**Q = Q**_{0} (1 - e - t/ Rc) where **Q = CV**_{0}

**VR = V**_{0} 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 = Q**_{0} e -t/RC' VR = V_{0} (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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**Capacitance Networks**

Charge distribution method

Star/ delta network

Wheatstone bridge cases

If

**C**

_{1}/C_{2}= C_{3}/C_{4}the eliminate C

_{5}.

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

**C**

_{1}and

**C**

_{2}charged to

**V**

_{1}and

**V**

_{2}are joined together then common potential is

**V**

_{common}= V_{1}C_{1}+ V_{2}C_{2}/C_{1}+ C_{2}= Q_{1}+ Q_{2}/C_{1}+ C_{2}Charge on capacitors after joining

**Q**

_{1}/Q_{2}= C_{1}/C_{2}

**Q**

_{1}= (Q_{1}+ Q_{2}) C_{1}/(C_{1}+ C_{2}Q_{2}= (Q_{1}+ Q_{2}) C_{2}/C_{1}+ C_{2}

Loss in energy when tow capacitors

**C**

_{1}and

**C**

_{2}charge to

**V**

_{1}and

**V**are joined together is

_{2}**?E = C**

_{1}C_{2}/2 (C_{1}+ C_{2}) (V_{1}- V_{2})^{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 V**

^{2}dC/dxGrowth transient or charging of a capacitor:- if Q is charge at any instant across the capacitor then

**Q = Q**where

_{0}(1 - e - t/ Rc)**Q = CV**

_{0}

**VR = V**

_{0}e^{-t}/RC**I = dQ/dt = Ra/RC e**

^{ - I}/RcTime constant τ is that during which a capacitor charges to 63% of its maximum value of charge.

Discharging of a capacitor (decay transient)

**Q = Q**

_{0}e -t/RC' VR = V_{0}(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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