Circuits-Physics?

Question

The circuit in the drawing shows two resistors, a C = 6.5 µF capacitor, and a V = 18 V battery. When the capacitor is fully charged, what is the magnitude q (in Coulombs) of the charge on one of its plates?

where R1 = 4 ohms and R2 = 2 ohms and are in series to each other and the capacitor is drawn in parallel w/ the 4 ohm resistance

Answer 1

OK, I hate Physics Webassign as well, and I feel your pain.

I'm not sure how to give you a direct explanation, so I'll give you the method and I guess you can derive the understanding from it..

CV / CR = I; where I = Current (in Ampere) and R = R1+R2 (since they are in series).

Once you solve for I (Current), you then multiply it with R1. For simplicity, lets call this answer 'X'

Now, multiply X by your given capacitance (IN FARADS!!! - 6.5 * 10^-6 I believe..)

and voila... that will give you the right answer.

Answer 2

Q = CV, but first you must find V, because C is given as 6.5uF.

V is found by finding the voltage of the resistor that the cap is in parallel with, because items in parallel experience the SAME voltage.

You stated that R1(4ohms) and R2(2ohms) were in series, and the cap is in parallel with 4 ohms. OK, so R1 and R2 make up a voltage divider, with R1 having the voltage we are interested in. So to find the voltage at R1 we must use a voltage division equation V1 = Vtotal(R1/(R1 + R2) = 18V(4ohms/(4ohms + 2 ohms) = 12V

Now we have what we need to find Q

Q = CV = 6.5uF * 12V = 78u coulombs (78 micro coulombs)

Answer 3

"Circuit/s" is used too many situations in UR question. ie =>what's "no longer for circuits" recommend on the top? that's perplexing given something of the words in this question Impedance refers to AC circuit/s and consists of components of: resistance capacitance inductance

Answer 4

Q = C*V