Question

to register when connected in series net resistance is 5 ohm and when they are connected in parallel net resistance is 1.2 ohm what are the registers​

Answer 1

R1=3ohm Then R2=2ohm

R1=2ohm then R2=3ohm

Explanation:

$R1 +R2= 5$

$R2 =5-R1$

$\frac{1}{R1} +\frac{1}{R2} =\frac{1}{1.2}$

On putting the value of R2 in above equation

$\frac{1}{R1} + \frac{1}{5-R1} = \frac{10}{12} \\ \\\frac{5-R1+R1}{R1(5-R1)}=\frac{10}{12}\\\\\frac{5}{R1(5-R1)}=\frac{10}{12} \\6=5R1-R1^{2} \\R1^{2}-5R1+6\\ R1-3 R1 -2R1+6\\R1(R1-3)-2(R1-3)\\(R1-3)=0\\R1=3\\(R1-2)=0\\R1=2$

When R1=3ohm then R2=5-3=2ohm

R1=2ohm Then R2=5-2=3ohm

Learn More :

• brainly.in/question/1433029

Answer 2

Explanation:

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