Help with this Circuit Problem!?

Question

Find the voltage drop across every resistor. Show your work

R1 =1
R2 = 3
R3 =5
R4 =1
R5 =7
R6 =2
R7 =4


---R5----R6----
I
---R2----(R3 and R4 in parallel)----
I
---R1---10V----R7----

This is the diagram of the circuit, I drew it as clear as possible I could. R5 and R6 is in series and below, R2 is in series with R3 and R4 that are in parallel, and below there are R1, Voltage, and R7 in series. Both left and right ends are connected vertically but I only drew on left side in thediagram. Hope you get the picture.. Thank you for your help!

Answer 1

Actully its an easy one i don't know what is the problems
find all the current in each resistor and then vind the voltage drop by multiply R * I

R5 + R6 =9
WE WILL CALL IT RA

R2+R3//R4 = 3.834
WE WILL CALL IT RB

RA//RB = (9*3.834)/(9+3.834)= 2.69
WE WILL CALL IT RC
THIS EQUATION IS APPLICABLE FOR TWO PARRLEL RESISTORS ONLY

RT = RC +R1+R7
RT = 2.69+1+7 = 10.69

iT = 10/10.69 = 0.935A

iB = 0.935 * 9 / (9+3.834) = 0.701A
CURRENT DIVIDER RULE
iA = iT - iB = 0.935 - 0.701 = 0.233A

now let us start working with the part A (R5 +R6)
Vd5 = iA*R5 = 0.233*7 = 1.631V
Vd6 = iA*R6 = 0.233*2 = 0.466V

now let us start working with the part B (R2+R3//R4)
R2 = 3
iB =i2 = 0.701A
Vd2 = R2*i2 = 3*0.701 = 2.103V

let us consider (R3//R4)
i3 = ib * R4 /(R3+R4)
current divider rule
i3 = 0.701 * 1 / (1 + 5) = 0.117A
Vd3 = i3*R3
Vd3 = 0.117*5 = 0.585V

i4 = iB - i3 = 0.701-0.117 = 0.584A
Vd4 = i4*R4
Vd4 = 0.584*1 = 0.585V

i1= i7 = it = 0.935A
Vd1 = 0.935 * 1 =0.935V
Vd7 = 0.935 * 4 = 3.74V

quick review for results

Vd1 = 0.935 * 1 =0.935V
Vd2 = 3*0.701 = 2.103V
Vd3 = 0.117*5 = 0.585V
Vd4 = 0.584*1 = 0.585V
notice Vd3 = Vd4 cause R3//R4
Vd5 = 0.233*7 = 1.631V
Vd6 = 0.233*2 = 0.466V
Vd7 = 0.935 * 4 = 3.74V

Answer 2

It looks like one parallel circuit that contains series resistors in each branch, with another parallel branch on its second branch. A Parallel within a Parallel.

Typical Parallel - Series Circuit.
1. Find the equivelent for the parallel (R3 & R4) then add this to the series R2 to get an equivelent resistor for this entire branch.
2. Add R5 & R6 to get an equivelent resistor for this branch
3. Add R1 & R7 to get an equivelent resistor for this branch

4. Figure out the parallel equivelent resistors

Answer 3

first find the resistance of the whole circuit.
There are three parallel paths.
The top is 7 + 2 = 9
the middle is 3 + (1/5 + 1/1) = 4.2
the bottom is 4

the whole circuit is then
(1 / 9) + (1 / 4.2) + (1 / 4) = 0.6

Now find the current
V = IR
10 = I * .6
I = 10 / .6 = 16.66