Solution

The 2k and the 4k are shorted out. Another way to think about it is that the 2k and 4k are in series, forming a 6k resistor. However, the 6k is in parallel with the wire (a wire is 0 ohms). Which is,

0 || 6k = 0*6k/(0+6k) = 0

Then we have the other two resistors in series, so

Req = 3k + 0 + 5k = 8k ohms

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© Calvin College, 2000
This page was written and is maintained by Steve VanderLeest.
It was last modified on 1 Mar 2000.