# Equivalent Resistance (Prep for Lab on DC circuits)

1. Feb 9, 2013

### TheOculus

1. The problem statement, all variables and given/known data
Use the circuit shown to answer the following questions. At time t = 0 the switch in the circuit is closed. Answer the following questions after the switch in the circuit is closed.

What is the equivalent resistance of the circuit given R1 = 5 , R2 = 5 , R3 = 4 , R4 = 5 , R5 = 1 and R6 = 2 and a 2 V battery?

2. Relevant equations
$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+...$ for parallel
$R_{eq}=R_1+R_2+...+R_n$ for series

3. The attempt at a solution

Alright, this didn't seem too complicated to me but all of my attempts have resulted in incorrect answers. The way I am seeing it is that $R_4$ and $R_6$ are parallel, and that the equivalent resistance of these two resistors are in series with the rest of the resistors. Here's my attempt:

$\frac{1}{R_{46}}=\frac{1}{5}+\frac{1}{2}=0.7Ω$ Now, the reciprocal of this is ≈1.43Ω which gives the equivalent resistance of resistors 4 and 6.
It seems like a simple matter here to add up this equivalent resistance to the rest of the resistors in series: 1.43+5+5+4+1=16.43Ω

Am I incorrect in my parallel/series diagnosis? I have tried several other options, which I won't type up here, but if my diagnosis is wrong I will offer up my other thoughts. Thanks in advance.

2. Feb 9, 2013

### Saitama

No, the equivalent of R_3, R_4 and R_5 is in parallel with R_6.

3. Feb 9, 2013

### TheOculus

Recalculating. Hadn't seen a resistor setup like that. I'm not sure I understand why they are in parallel. Is it because there is no wire between, i.e. they are connected to each other?

Edit: Ok, that worked like a charm. Thanks.