Math, asked by ghost48, 5 months ago

17. Read the following and answer any four questions from 17 0 to 12. (V)
Radhika went to an electronic shop to get her radio repaired. The electrician
required resistances of 3 N and 14 N to repair the radio set. He had a large number
of 4 N resistors. So, he made attempts but could not get the correct combination of 3
N and 14 O resistances. Radhika has studied the combination of resistors and helped
the electrician to arrange the 4A resistors to obtain the required resistances.
O low would Radhika have arranged the 4 A resistors to obtain the 3 N resistance?​

Answers

Answered by amitnrw
1

Given : required resistances of 3 N  using 4N

To Find :  How to achieve

Solution:

Take 3 Resistor

4N resistors ranged in series will be multiple of 4

Required resistance = 3N  =  3 * 4N / 4  = 12N / 4

Hence if we add 4  ,  12N resistors in parallel , net resistance would be

= 1/(1/12N + 1/12N  + 1/12N  + 1/12N)

= 12N/4

= 3N  

12 N resistance can be achieved by adding  4N resistance 3 times in series .

So to achieve 3N resistance from  4N resistors

12 resistors of 4N are required  which will arranged in way

4  Parallel combination of  ( 3 series resistors of 4N)

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Answered by nareshsaini77117
0

Step-by-step explanation:

Given : required resistances of 3 N  using 4N

To Find :  How to achieve

Solution:

Take 3 Resistor

4N resistors ranged in series will be multiple of 4

Required resistance = 3N  =  3 * 4N / 4  = 12N / 4

Hence if we add 4  ,  12N resistors in parallel , net resistance would be

= 1/(1/12N + 1/12N  + 1/12N  + 1/12N)

= 12N/4

= 3N  

12 N resistance can be achieved by adding  4N resistance 3 times in series .

So to achieve 3N resistance from  4N resistors

12 resistors of 4N are required  which will arranged in way

4  Parallel combination of  ( 3 series resistors of 4N)

Learn More:

Two resistors of resistance 1.0 Ω and 2.0 Ω connected in parallel are ...

brainly.in/question/18350094

“series combination” and “parallel combination”

brainly.in/question/11390122

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