Kindly explain how to solve the following network. The answer is 6 ohms and independent of R2. Can anyone explain that?
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Remove R2 and you'll see two parallel sets of 6 ohms (3 + 3) in parallel. If you have two resistors (6 ohms each), I can assure you that the equivalent resistance is NOT 6 ohms for the full circuit.Kindly explain how to solve the following network. The answer is 6 ohms and independent of R2. Can anyone explain that?
That's what I'm seeing. Just how the TS knows it's 6Ω is beyond me. Maybe it is. I've always been good at math. There's NO equation I can't screw up!3 ohms...
Thanks for the reply. Yes, I just misquoted it. The answer is 3 ohms. Really sorry for that. But can you explain how do we arrive at 3 ohms.Remove R2 and you'll see two parallel sets of 6 ohms (3 + 3) in parallel. If you have two resistors (6 ohms each), I can assure you that the equivalent resistance is NOT 6 ohms for the full circuit.
The answer is 3 ohms. Sorry for misquoting.Kindly explain how to solve the following network. The answer is 6 ohms and independent of R2. Can anyone explain that?
Yes, I can see that there is no potential difference across 400 ohms. So we eliminate it from the circuit and solve. Is that right?Can you solve it now ? ? ?View attachment 131895
Yes, I am 100% clear. Thanks.As I gave you the hint before, the 400Ω resistor has no current. Therefore no voltage. No voltage, no current no apparent resistance. In other words the 400Ω resistor is just a goldbrick circuit. It does nothing to the values.
SO: You have two 3Ω resistors in series  effectively creating a 6Ω resistor. You have a second set of resistors constituting another 6Ω resistor. Two 6Ω resistors in parallel equate to 3Ω. Do you understand how we come up with 3Ω?