Hi guys just looking for some help really

friswell_38 · Aug 16, 2011

Hi i am ex army currently retraining i have completed my level 2 2330 and nvq just starting my level 3 and doing it as a distance learning course while working so not the easiest way to learn sometimes but no choice i have commitments.
anyways im stuck on a question while doing some revision work .

Q. how much charge will develop in the capacitor ?

current - 200 mA into a capacitor for 40 mS

any help would be much appriciated

vernam616 · Aug 16, 2011

Charge is equal to current x time
0.2 x 0.04 = 0.008C or 8mC

hope this helps

friswell_38 · Aug 16, 2011

Thats awesome thankyou for your help

friswell_38 · Aug 18, 2011

hey me again just one more question if you have the time please .
a capacitor 95 uF is connected in parralell with a 20 ohm resistor on a 230 v 50 hz supply.

What is the dissipated power? how would i calculate this? Is dissipated power the same as potential difference?

MarkieSparkie · Aug 19, 2011

Definitions
Power Dissipated is the energy consumed doing useful work (eg. driving a motor) or lost in the circuit (eg. in heat), measured in Watts.
Potential Difference is the electromotive force which drives a current between two points on a circuit, measured in Volts.
Solution to Problem
Capacitive Reactance:
X[SUB]c[/SUB]=1/2PifC
=1/6.28x50x95E-6
=33.5 Ohms
Resistor Current:
I[SUB]r[/SUB]=V/R
=230/20
=11.5 Amperes
Capacitor Current:
I[SUB]c[/SUB]=V/X[SUB]c
[/SUB]=230/33.5
=6.87 Amperes
Supply Current:
=SQR(I[SUB]r[SUP]2[/SUP][/SUB]+I[SUB]c[SUP]2[/SUP][/SUB])
=SQR(11.5[SUP]2[/SUP]+6.87[SUP]2[/SUP] [SUB])[/SUB]
=13.4Amperes
Supply Current Leads Voltage by:
Phi=tan[SUP]-1[/SUP](I[SUB]c[/SUB]/I[SUB]r[/SUB])
=tan[SUP]-1 [/SUP](6.87/11.5)
=30.85 Degrees
Power Dissipated in Circuit:
P=VI cosPhi
=230x13.4xcos30.85
=2646 Watts