a collection of two 2ohms , one 3 ohm and one 4 ohm resistors are provided. connect them in a suitable order to get an effective resistance of 2 ohm. support your answer with suitable calculations.
Answers
Answer:
Given: R1=2ohm, R2=3 ohm, R3=6ohm
Formula: R1=R11+R21+R31=21+31+61
∴R=63+2+1=66=1ohm.
Explanation:
On combining two parallelly connected 2 ohms resistance to 3 ohms resistance in series and then connecting this system parallelly to 2 ohms resistance, we get the resultant resistance of 2 ohms.
Combination of resistors:
We are given four resistors, each of resistance 2 ohms, 3 ohms, and 4 ohms such that their equivalent resistance becomes 2 ohms.
We can connect the elements either in series or in parallel. The resistors connected in parallel will have lower equivalent resistance whereas the ones connected in series will have higher resultant equivalence.
Here, let's connect the two 2 ohms resistance in parallel:
1÷R'= 1÷R1 +1÷R2
R'= R1× R2÷ R1+ R2
R'= 2×2÷ 2+ 2
R'= 4÷4= 1
Now, this system of combination should be linked to the 3 ohms resistance in series:
R''= R1+ R2
R''= 1 + 3
R''= 4 ohms
Now, the above system should be linked to the 4 ohm resistance in parallel:
= R1× R2÷ R1+ R2
= 4 ×4÷ 4 +4
= 16÷ 8
=2 ohms
Result:
Thus, on combining two parallelly connected 2 ohms resistance to 3 ohms resistance in series and then connecting this system parallelly to 2 ohms resistance, we get the resultant resistance of 2 ohms.
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