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## Homework Statement

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R1 = 42 Ω

R2 = 124 Ω

R3 = 119 Ω

R4 = 105 Ω

Rx = ?

I4 = O A

## Homework Equations

R = IV

V = V1 = V2 for parallel Resistors

Iequiv = I1 +I2 for parallel Resistors

Vequiv = V1+V2 for series Resistors

Iequiv = I1= I2 for series Resistors

## The Attempt at a Solution

I have to find I1, the current that flows through resistor 1. But I am unsure of how to start. The first thing I tried was getting the combined resistance of R1 and R2 (R12):

(1) 1/R12 = 1/ R1 +1/R2

Since they are parallel I applied the rule that Resistors in parallel have equal voltages:

(2) Vequiv = V1 = V2

but I wasn't sure that would work unless I packed all the resistors together first (and since I don't have a value for Rx yet I am unsure how to proceed in this manner)

So, instead I tried using a ratio. Using V = IR and using the Voltage relationship for parallel resistors I tried:

(3) I1/I2 = R2/R1

I got a value for I1 in terms of I2 and set up an equation using the rule Iequiv = I1 +I2

From this I got

(4) I1/I2 = 62/21 (5) I1 = 62/21(I2)

62/21(I2)+I2 = 12V (voltage of the battery)

solving (4) for I2 I got 252/63

subbing back into the eq (5) I got that I1 = 744/83 but this was wrong.

I think I am misunderstanding how the relationships (Vequiv = V1 = V2 )for parallel and (Iequiv = I1 = I2) for series resistors works. Can someone help clarify and/or explain where I am going wrong in my logic?

** I apologize for formatting. I tried to use LATEX code but I couldn't get it to work properly and with finals next week I simply haven't the time to spend learning it for this question. Please forgive me! :)

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