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In the formula 1/R=1/R1
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14 Jul 2021, 05:05
14 Jul 2021, 05:05
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In the formula \(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\), if \(R=4 \) and \(R_1=10\), what is the value of \(R_2 \)?
Express your value as a fraction
Express your value as a fraction
Show: :: OA
\(\frac{20}{3}\)
Re: In the formula 1/R=1/R1
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14 Jul 2021, 05:30
14 Jul 2021, 05:30
1
Solution:
\(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\)
\(\frac{1}{R}- \frac{1}{R_1}= \frac{1}{R_2}\)
\(\frac{1}{4 }- \frac{1}{10}= \frac{1}{R_2}\)
\(\frac{6}{40}= \frac{1}{R_2}\)
\(\frac{3}{20}= \frac{1}{R_2}\)
\(R_2=\frac{ 20}{3}\)
Answer: \(\frac{ 20}{3}\)
Hope this helps!
\(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\)
\(\frac{1}{R}- \frac{1}{R_1}= \frac{1}{R_2}\)
\(\frac{1}{4 }- \frac{1}{10}= \frac{1}{R_2}\)
\(\frac{6}{40}= \frac{1}{R_2}\)
\(\frac{3}{20}= \frac{1}{R_2}\)
\(R_2=\frac{ 20}{3}\)
Answer: \(\frac{ 20}{3}\)
Hope this helps!
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In the formula 1/R=1/R1
[#permalink]
14 Jul 2021, 05:55
14 Jul 2021, 05:55
1
\(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\)
Substituting the value of \(R=4 \) and \(R_1=10\) we get
\(\frac{1}{4}=\frac{1}{10}+\frac{1}{R_2}\)
=> \(\frac{1}{R_2}\) = \(\frac{1}{4} - \frac{1}{10}\)
We know that LCM(4,10) = 20
2 | 4 , 10
-----------
1 | 2 , 5
=> LCM(4,10) = 2*2*5 = 20
=> \(\frac{1}{R_2}\) = \(\frac{5}{20} - \frac{2}{20}\) = \(\frac{5-2}{20}\) = \(\frac{3}{20}\)
=> \(R_2\) = \(\frac{20}{3}\)
Hope it helps!
To learn How to find LCM and GCD using multiple methods, watch the following video
Substituting the value of \(R=4 \) and \(R_1=10\) we get
\(\frac{1}{4}=\frac{1}{10}+\frac{1}{R_2}\)
=> \(\frac{1}{R_2}\) = \(\frac{1}{4} - \frac{1}{10}\)
We know that LCM(4,10) = 20
2 | 4 , 10
-----------
1 | 2 , 5
=> LCM(4,10) = 2*2*5 = 20
=> \(\frac{1}{R_2}\) = \(\frac{5}{20} - \frac{2}{20}\) = \(\frac{5-2}{20}\) = \(\frac{3}{20}\)
=> \(R_2\) = \(\frac{20}{3}\)
Hope it helps!
To learn How to find LCM and GCD using multiple methods, watch the following video
