Posted by **Elizabeth** on Thursday, February 21, 2013 at 2:36am.

Two resistances, R1 and R2, are connected in series across a 15-V battery. The current increases by 0.050 A when R2 is removed, leaving R1 connected across the battery. However, the current increases by just 0.025 A when R1 is removed, leaving R2 connected across the battery.

Find R1 and R2

- Physics -
**JJ**, Thursday, February 21, 2013 at 1:05pm
V= IR

in this case V = I(R1+R2)

so we have 15 = IR1 + IR2

(15 - IR1)/I = R2 we will use this later in our subsitution.

We also know that...

15 =(I+.05)R1 and 15 = (I+.025)R2

since both equal 15, they must equal each other:

(I+.05)R1 = (I+.025)R2

Use the substitution for R2 from above and you will be able to solve for R1. by eliminating the I.

## Answer this Question

## Related Questions

- PHYSICS - When the resistors are connected in series to a 12.0-V battery, the ...
- Physics - Two resistors have resistances R1 and R2. When the resistors are ...
- Science - When 2 resistors, R1 and R2 are connected in series across a 6V ...
- pHYSICS - Two resistors have resistances R(smaller) and R(larger), where R(...
- PHYSICS 2 - Two resistors have resistances R(smaller) and R(larger), where R(...
- Physics - 1)Several resistors are connected in series. If a battery provides ...
- Physics - A 2.0-ohm resistor is connected in series with a 20.0-V battery and a ...
- Physics - A battery charger is connected to a dead battery and delivers a ...
- Physics - A 79.4-Ω and a 47.7-Ω resistor are connected in parallel. ...
- AP PHYSICS - A 3.0-uF capacitor and a 4.0-uF capacitor are connected in series ...

More Related Questions