How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?

Q. How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?

 

Answer:-



There are three resistors of resistances 2 Ω, 3 Ω, and 6 Ω respectively.
(a) If we connect the given three resistors in the following way then we get net resistance as 4Ω
chapter12o
Here, 6 Ω and 3 Ω resistors are connected in parallel.
Therefore, their equivalent resistance will be given by
chapter12p
⇒R=2 Ω
This equivalent resistor of resistance 2Ω is connected to a 2Ω resistor in series.
Therefore, equivalent resistance of the circuit = 2Ω + 2Ω =4Ω
Hence, the total resistance of the circuit is 4Ω.
(b) If we connect the given three resistors in the following way then we get net resistance as 1Ω. Here all the resistors are connected in parallel
chapter12q
∴ their equivalent resistance will be given as
chapter12or
R=1Ω
Therefore, the net equivalent resistance of the circuit is 1Ω.


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