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The given equation mentions that there are three resistors each with resistances of 2 Ω, 3 Ω, and 6 Ω respectively.
Assume that,
- R1 = 2 Ω
- R2 = 3 Ω
- R3 = 6 Ω
Read More: Kirchhoff’s Laws
- Total resistance 4 ohm (Ω)
In order to acquire a total resistance of 4 Ω from the three resistors, attach the resistors of 3Ω and 6Ω parallelly to obtain a total resistance of 2Ω (that is lesser than the lowest individual resistance). Now, when R2 and R3 are parallelly connected with R1 in series, we can obtain:
⇒ 1/R = 1/R2 + 1/R3
⇒ 1/R = 1/3 + 1/6
⇒ 1/R = 1/2
⇒ R = 2 Ω
⇒ Thus, the Resistance in series = R + R1 = 2 + 2 = 4 Ω
Therefore, the total resistance is 4 Ω.
- To obtain a total resistance of 1Ω from three resistors of 2Ω, 3Ω and 6Q, all three resistors should be connected in parallel.
Considering that R1, R2, and R3 are parallelly connected:
⇒ 1/R = 1/R1 + 1/R2 + 1/R3
⇒ 1/R = 1/2 + 1/3 + 1/6
⇒ 1/R = 1 Ω
Thus, the total resistance is 1 Ω.
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