[tex]\normalsize\sf\red{Find \: \: the \: \: equivalent \: resistance \: between } \\ \normalsize\sf\red{points \: \: A \: \: and \: \: B. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }[/tex]
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GiveN :-
To FinD :-
CalculationS :-
We will simply the circuit step by step to obtain the equivalent resistance . The given circuit looks a bit complex so let's transform it. Refer to the attachment.
Now we see that the resistances of 12Ω , 4Ω and 6Ω are connected in parallel .
• So the net resistance will be ,
Finally we see that the resistances of 2Ω , 2Ω and 4Ω are in series . So the final net resistance will be ,
Answer:
tum ne to bola tha ki wo meri id waha se points le lo chahiye to yad aaya