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Adding a resistance in parallel to another resistance will reduce the equivalent resistance.

Parallel Equation:

R1 = 10Ω

R2 = 20Ω

Req = R1R2/(R1+R2) = (1/R1+1/R2)-1

Req = 10•20/(10+20) = 6.67Ω

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10y ago
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Hilmar Zonneveld

Lvl 7
4y ago

In that case, the resulting (equivalent) resistance will be less than any of the individual resistances.

The equivalent resistance is calculated by the formula:

1 / R = 1 / R1 + 1 / R2 + 1 / R3 ...

Where "R" is the equivalent resistance you want to calculate, and R1, R2, etc. are the resistance values of the individual resistors.

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Q: What happens to the resistance of the circuit when you add resistances in parallel?
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