I have a parallel circuit with 7.5 ohms as total resistance,75 volts, with 2 resistors. R1=10ohms,R2 is unknown. How can I find the value of R2?

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    \$\begingroup\$ Show your circuit and your work. \$\endgroup\$ – DerStrom8 Feb 19 '17 at 0:57
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    \$\begingroup\$ I'm voting to close this question as off-topic because it appears to be a homework question with no prior attempts shown, and the OP couldn't even be bothered to provide a schematic. \$\endgroup\$ – DerStrom8 Feb 19 '17 at 0:58

Since \$R_{tot}=\frac{R_1\cdot R_2}{R_1+R_2}\$ it follows from simple algebra that \$R_2=\frac{R_1\cdot R_{tot}}{R_1- R_{tot}}\$. In this case, that means \$R_2=\frac{10\:\Omega\cdot 7.5\:\Omega}{10\:\Omega- 7.5\:\Omega}=30\:\Omega\$. The voltage you mentioned isn't required.

I'd recommend that you commit both parallel resistance formulas to memory, as well. But use some algebra and prove what I said is right. You may need the practice.

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    \$\begingroup\$ Maybe a bit easier to do 1/R2 = 1/Rt - 1/R1 = 1/7.5-1/10 \$\endgroup\$ – Spehro Pefhany Feb 19 '17 at 1:10
  • \$\begingroup\$ You really ought not give direct answers. For all we know you could have just given the OP an answer on a test that he or she was supposed to do on their own \$\endgroup\$ – DerStrom8 Feb 19 '17 at 2:00
  • \$\begingroup\$ @DerStrom8 I'm rarely focused on the OP, except as a guide about what they see. It helps other people. I use the questions as a foil of sorts. When I'm in the mood and the questioner doesn't otherwise make a simple answer easy to misdirect. In this case, there's little harm possible I think ... even to the OP. I feel fine about it. \$\endgroup\$ – jonk Feb 19 '17 at 4:38
  • \$\begingroup\$ @SpehroPefhany Yeah, but then you have to do a 1/x again. Anyway, it's very helpful to just memorize the fact that putting them in parallel is a + sign in the denominator and that extracting one or the other uses a - sign. But sure. Any way that works for someone is all good. \$\endgroup\$ – jonk Feb 19 '17 at 4:40
  • \$\begingroup\$ @jonk Re. Spehro's comment, the formula you provide only works when you have two parallel resistors. It may be misleading if the reader has 3 or more resistors in parallel. You cannot just multiply by the third resistance in the numerator and add it in the denominator, you really must use the formula that Spehro provided. \$\endgroup\$ – DerStrom8 Feb 19 '17 at 14:22

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