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## Homework Statement

For 1 and 2 : Find the equivalent resistance for the resistor network provided

## Homework Equations

For resistor in series : Req = R[itex]_{1}[/itex]+R[itex]_{2}[/itex]+...R[itex]_{n}[/itex]

For resistors in parallel : Req = (1/R[itex]_{1}[/itex]+1/R[itex]_{2}[/itex]+... 1/R[itex]_{n}[/itex])[itex]^{-1}[/itex]

## The Attempt at a Solution

For

**question 1**:

I have considered the two paths shown :

1st path (blue) : total resistance = 2 ohm

2nd path (orange) : total resistance = 2 ohm

so since these two paths are in parallel , Req = 1 ohm

But answer provided = 2/3 ohm

For

**Question 2**:

Since no current passes through the middle wire, upper path = 10 ohm

lower path = 10 ohm

So, total resistance = 5 ohm

However , the answer used the method of (1/4+1/8)^-1 + (1/2+1/6)^-1 = 4.17 ohm

But I don't understand this part, if there is current flowing in the middle wire, there would be one current flowing down and one current flowing up. Won't they cancel out each other? Thanks !!