Question

What is (a) highest, and (b) lowest, resistance which can be obtained by combining four resistors having the following resistances?
4 Ω, 8 Ω, 12 Ω, 24 Ω; What is (a) highest, and (b) lowest, resistance which can be obtained by combining four resistors having the following resistances?
4 Ω, 8 Ω, 12 Ω, 24 Ω

Answer 1

Explanation:

Given resistances R1 =4 Ω, R2=8 Ω, R3=12Ω, and R4=24Ω.

(a) If these coils are connected in series, then the equivalent resistance will

be the highest, as R=R1+R2+R3+R4

∴ R=4+8+12+24 = 24Ω

(b) If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by

⇒R=2 Ω

∴ 2 Ω is the lowest total resotar

Answer 2

(a) The highest resistance is 48 Ω.

(b)The lowest resistance is 2 Ω.

Explanation:

(a) All the resistors must be connected in series to get the optimum resistance.

Resistance is provided in a series arrangement by

$R = R_1 +R_2 + R_3+ R_4$

R= 4 Ω+ 8 Ω+ 12 Ω+ 24 Ω  

R= 48 Ω

The highest resistance is therefore 48 Ω.

(b) All the resistors must be wired in parallel to obtain the lowest resistor.

Resistance is given in a parallel arrangement through:

$\frac{1}{R} =\frac{1}{R_1 } +\frac{1}{R_2} +\frac{1}{R_3} +\frac{1}{R_4 }$

Here $R_1 = 4 \Omega, R_2 = 8 \Omega, R_3 = 12 \Omega, R_4 = 24 \Omega$

$\frac{1}{R} =\frac{1}{4} +\frac{1}{8} +\frac{1}{2} +\frac{1}{24}$

$\frac{1}{R} =\frac{(6+3+2+1)}{24}$

$\frac{1}{R} =\frac{12}{24}$

R=2 Ω

Therefore, the arrangement's lowest resistance is 2 Ω.

You can get all the answers of Chapter 1 in the link below:    

https://brainly.in/question/14961133

   

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