Question

What is the highest and lowest resistance that can be obtained by 4 resistors of resistance 4Ω, 8Ω, 12Ω and 24Ω?

Also tell how will you connect these resistors.

Answer 1

Answer:

Ans: Given resistances R1 =4 Ω, R2=8 Ω, R3=12Ω, and R4=24Ω. ∴ 2 Ω is the lowest total resistance.

Answer 2

Required Answer:-

To obtain the highest resistance, you have to connect all the resistors in series connection.

Resistance in series can be expressed as::

⇒ R=R₁+R₂+R₃...

By substituting values of resistors:

⇒ R=4Ω+8Ω+12Ω+24Ω

⇒ R=48Ω

So the highest resistance which can be obtained by given resistors is 48Ω.

__________________________

To obtain the lowest resistance, we have to connect all the resistors in parallel connection.

Resistors in parallel can be expressed as::

⇒ 1/R=1/R₁+1/R₂+1/R3...

By substituting values of resistors:

⇒ 1/R=1/4+1/8+1/12+1/24

⇒ 1/R=(6+3+2+1)/24

⇒ 1/R=12/24

⇒ 1/R=1/2

By cross multiplication, we get:

⇒ R=2Ω

So lowest resistance which can be obtained by given resistors is 2Ω.