Calculating current through each resistor and potential diffrence across?

Question

http://i102.photobucket.com/albums/m93/XedLos/Untitled.gif
R1 = 8.00 , R2 = 11.0 and R3 = 2.0
(a) Determine the current in each resistor.
(b) Determine the potential difference across each resistor.

Answer 1

There are many approaches to arriving to an approximate steady state solution, so here is one general solution:

Step One: Determine useful values and formula.

Vs = 12V
R1 = 8 (ohms)
R2 = 11 (ohms)
R3 = 2 (ohms)

V = Volts = measure of potential energy along e-fields
I = Amperes = measure of flux of square coulombs/second
Kirchoff's Law: dV/dI = R; dV = R*dI
This simplifies for linear circuits to the commonly used relationship of Kirchoff's Law (simplified): V = R x I
Parallel resistors have an equivalent shared resistance that is perceived to be as follows:
(1 / Requivalent) = sum of reciprocal resistances =
(1 / R1) + (1 / R2) ... + (1 / R(n -1)) + (1 / Rn)
Series resistors have an equivalent total resistance that is seen as: Rtotal = sum of resistances = Ra + Rb +... + Rfinal
Voltage division: for R1, R2, Vs - V1,2 = Va
Va = Vs x [(R1)/(R1 + R2)]
Current division: for R1, R2, Itotal
I through R1 = Ir1 = Itotal x [(R2)/(R1 + R2)]
I through R2 = Ir2 = Itotal x [(R1)/(R1 + R2)]

Solution Steps:
Find total current Itotal = Vs/Rtotal

Vs = 12; Rtotal (in ohms) = [1 / ((1/R1) + (1/R2))] + R3
Rtotal = [1 / ((1/8) + (1/11))] + 2
= 4.63 (rounded to nearest hundredth of an ohm) + 2
= 6.63 ohms = Rtotal
Vs/Rtotal = 12/6.63 = 1.81 amperes = 1.81A

Va = Vs x [(R1equivalent) / (R1equivalent + R2)] =
12V x [4.63/6.63] = drop over R1 & R2 = 8.38V
Vb = drop over R3 = 12V - 8.38V = 3.62V

Ir3 = Itotal = 1.81A
Ir1 = Itotal x [R2/(R1 + R2)] = 1.81A x [11/19] = 1.05A (rounded up)
Ir2 = Itotal - Ir1 = 1.81A - 1.05A = 0.76A

To see if I did this correctly, and to see if you understand the process of getting to the solution, please check my work for yourself.

That means plug values in and see if they match up and follow the rules for a passive linear circuit.

Remember, the voltage minus the voltage drops around the complete circuit (loop) must equate to zero.
Remember, the current coming out of a node must equal zero if no energy is produced at that node.
That means: What goes in, must come out. That means that if a current splits from one flow to more than one, the addition of all the flows must equal the source.

Okay, well, there you are. I hope I helped, that this help you learn how to do a basic circuitry solution, and I hope I did not just help you to cheat on homework/exam.

Good Day!

~X

Answer 2

so, for the electric circuit that I saw in the picture this is what I do:
E = 12 V
R1 = 8 Ohms
R2 = 11 Ohms
R3 = 2 Ohms
- I found the equivalent electric resistance for all the circuit
Re = (R1 x R2)/(R1+R2) + R3 = 88/19 + 2 = 126/19 Ohms
because R1 and R2 are in parallel and the resulting resistance is serial with R3
- I found the current that passes through resistance R3
i3 = E/Re = (12 x 19)/126 = 1,809 A
- the voltage drop on R3 is
U3 = i3 x R3 = 1,809 x 2 = 3,618
- the current i3 can be written as
i3 = i2 + i1 Kirchhoff
- the voltage drop on resistance R1 and R2 is the same (paralel circuit)
U1=U2= E - U3 = 12 - 3,618 =8,382 V
- so, the currents i1 and i2 are
i1 = U1/R1 = 8,382/8 = 1,047 A
i2 = U2/R2 = 8,382/11 = 0,762 A
- and, to verify the results
i1 + i2 = 1,047 + 0,762 = 1,809 A = i3

Answer 3

- Resistors #1 and #2 are in parallel, so you can resolve them into equivalent resistance using the formula 1/R = 1/R1 + 1/R2.

- Therefore, 1/R = 1/8 ohms + 1/11 ohms => R = 4.63 ohms. Now you have a circuit "two resistors", both in series. One is 4.63 ohms and the other is 2.0 ohms.

- Now you can use Ohm's Law to determine the current and voltage across the resistors (V = IR).

If you need any more help, send me a message.

Answer 4

k, u need to tell me the emf, or the the voltage before we can calculate the current, and tell me if they are parallel or series