Please Help
Please Help
Was wondering if anyone could possibly help with this very simple question which I keep going round and round with but the sums do not seem to add up in terms of kechoffs V Law
Input V = 6V
Capacitor 1 = 2 Micro Farads
Capacitor 2 = 2 Capacitors in Parallel 1 @ 0.4 Micro Farads and 1 @ 0.6 Micro Farads
a) What is the total Charge into Network
b) What is the Charge associated with the 0.4 Micro Farad Capacitor
c) Calculate energy stored on 0.6 Micro Farad Capacitor
d) Calculate total Energy Stored.
Any help would be much apprecciated
Input V = 6V
Capacitor 1 = 2 Micro Farads
Capacitor 2 = 2 Capacitors in Parallel 1 @ 0.4 Micro Farads and 1 @ 0.6 Micro Farads
a) What is the total Charge into Network
b) What is the Charge associated with the 0.4 Micro Farad Capacitor
c) Calculate energy stored on 0.6 Micro Farad Capacitor
d) Calculate total Energy Stored.
Any help would be much apprecciated
Re: Please Help
Is this the circuit you want to analyze?:
- Attachments
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- Capacitor circuit
- capacitors.jpg (54.4 KiB) Viewed 7129 times
Re: Please Help
That is exactaly the circuit I was wanting to analyze. Thankyou very much for the outstanding schematic but I realised I was being stupid and using the resistor formula for capacitors so all sorted now. Thanks very much for your help though.
Re: Please Help
You're welcome - I'm glad you found a solution 
Re: Please Help
I follow above circuit,I didn't get solution.
Any other soultions.
Any other soultions.
Re: Please Help
a) Q = C * V (charge [C] = capacitance [F] * voltage [V])
Qtotal = Ctotal * Vin
Ctotal = 1 / ([1 / C1] + [1 / Cparallel])
Cparallel = C21 + C22 = 0.4 u + 0.6 u = 1 uF
Ctotal = 1 / ([1 / 2 u] + [1 / 1 u]) = 1 / (500 k + 1 M) = 1 / 1.5 M = 0.6667 uF
Qtotal = 0.6667 u * 6 = 4.000 u ~ 4 uC
b) c) ?
d) W = 0.5 * C * V² (energy [J] = 0.5 * capacitance [F] * voltage [V]²)
Wtotal = 0.5 * Ctotal * Vin² = 0.5 * 0.6667 u * 6² = 12.00 u ~ 12 uJ
Qtotal = Ctotal * Vin
Ctotal = 1 / ([1 / C1] + [1 / Cparallel])
Cparallel = C21 + C22 = 0.4 u + 0.6 u = 1 uF
Ctotal = 1 / ([1 / 2 u] + [1 / 1 u]) = 1 / (500 k + 1 M) = 1 / 1.5 M = 0.6667 uF
Qtotal = 0.6667 u * 6 = 4.000 u ~ 4 uC
b) c) ?
d) W = 0.5 * C * V² (energy [J] = 0.5 * capacitance [F] * voltage [V]²)
Wtotal = 0.5 * Ctotal * Vin² = 0.5 * 0.6667 u * 6² = 12.00 u ~ 12 uJ