Physics: Direct Current Circuits, Help Would Be So Appreciated?

Question

Hi everyone. I've been trying to do my WebAssign for AP Physics for over a few hours now, and I've done every question except one (that has two parts). Any help would be wonderful. I have a test on these types of questions tomorrow, and I would greatly appreciate any help from anyone.

The question(s) is/are as follows:

R = 15 ohms
(a) Find the current in the 15.0 ohm resistor.
(b) Find the potential difference between points a and b.

This is the link to a picture of the circuit: http://i31.photobucket.com/albums/c395/thisusernameisthebes/p28-09alt.gif

I would so very much appreciate any help from anyone. It seems like a simple question, but is causing me so much grief. Thank you so much everyone!

Answer 1

If R = 15 ohms then R+5ohms =20ohms
Recognize that the 10 ohm, the 5 ohm, and the R+5 ohm are a parallel circuit.
1/10+1/5+1/20=1/total
total=1/(7/20)=2.86ohms
This is in series with the battery's 10 ohms.
total resistance = 12.86 ohms
Voltage across the battery's 10 ohms=
10/12.86*25=19.44 Volts
Voltage across a and b = 25-19.44=5.56Volts
Cross-check 2.86/12.86*25 = 5.56 volts
Ohm's law E= I * R
5.56 = I * 20 (remember you need the total resistance of this branch, not just the R=15)
I = 5.56/20 = 0.278 Amps
(a) 0.278 Amperes
(b) 5.56 Volts

Answer 2

ok, so which you have the capacitors in sequence in the time of a ninety 5 V battery. Say that Capacitor C1 could have a voltage drop in the time of it of V1 and Capacitor C2 could have a voltage drop in the time of it of V2. V1 + V2 = ninety 5 V. Now all of us be responsive to that the charge on C1 will equivalent the charge on C2. so as which potential Q1 = Q2. We additionally be responsive to that Q = C*V. So if Q1 = Q2 then we are in a position to rewrite this as C1*V1 = C2*V2 permit's remedy for V2 and substitute that expression into the voltage equation: if C1*V1 = C2*V2 then V2 = C1*V1/C2 and V1 + (C1*V1/C2) = ninety 5 V. V1* (a million + (C1/C2)) = ninety 5 V. V1 = ninety 5/(a million + (C1/C2)) volts all of us be responsive to C1 = 4 microF and C2 = 2 microF so V1 = ninety 5/(3) volts and because C = Q * V, then Q = C/V Q1 = 4 * 10^-6 Farads / (ninety 5/3) volts = a million.263 * 10^-7 C. Q2 is the comparable. Now connect them in parallel to a minimum of one yet another. The voltage drop in the time of the two capacitors could desire to now be equivalent and the charge will redistribute to realize that. In parallel V1 = V2 so Q1/C1 = Q2/C2. We additionally be responsive to that the completed charge around the capacitors could desire to stay the comparable because of the fact there is not any path to leak charge off to any region. So Q = Q1 + Q2 = 2 * a million.26 * 10^-7 C = 2.526 * 10^-7 From the voltage equation we are in a position to remedy for between the Qs in terms of the different and substitute it into the charge equation. Q1 = Q2 *(C1/C2) So the charge equation will become... Q2* (C1/C2) + Q2 = 2.526 * 10^-7 Q2 * ((C1/C2) + a million) = 2.526 * 10^-7 Q2 * (3) = 2.526 * 10^-7 Q2 = 8.421 * 10^-8 C you are able to then remedy for Q1.

Answer 3

I get 50/180 amps for part a and 50/9 Volts for part b.