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For the given circuit, I was asked to calculate the eqv. resistance: enter image description here

But, I'm not sure what does grounding some of the resistors do? In which way does it change the calculation or is there no difference whatsoever for the given?

(there is no need to solve the question, just an explanation would suffice, thanks i advance)

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    \$\begingroup\$ Hello Aleks, simply wire all of the ground terminals together. That's what the ground symbol implies. (Usually, but not always, this is also connected to Earth ground.) \$\endgroup\$ – rdtsc Feb 22 '16 at 19:56
  • \$\begingroup\$ Thanks I just solved it, it's R/2 we get a Wheatstone bridge after wiring everything together, and the resistor in the middle has no effect, so it becomes really simple. (I cannot upvote your answer sorry not enough rep) \$\endgroup\$ – AleksXPO Feb 22 '16 at 19:59
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The ground symbol is just a shortcut. It's the most common connection in any schematic, which would become visually cluttered if all the connections were made explicitly. Just imagine there's a wire (a perfect wire with zero resistance/inductance, not the case in the real world) drawn between all the ground symbols and go from there.

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Those grounds are just wired connections!

One can redraw the circuit as

schematic

simulate this circuit – Schematic created using CircuitLab

Solve this by Star-Delta / Delta-Y conversion. There are 4 stars/Ys in the circuit. Looked differently there are 3 deltas. Convert one of them like R1-R3-R5 delta into a star/Y, you can reduce the following easily into series and parallel combinations.

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