# Finding voltage difference in a resistance temperature device

I would like to know if I am on the right track on this question The way I approached this question is as follows.

$$R = 500[1+0.005(100-30)]\\R = 675\Omega \\$$ Then I find $$R_{eq} = 540.23 \Omega$$ I found the total current going into the branches $$I = \frac{10}{R_{eq}} = 18.5 mA$$

I applied the current division and found the current going through RT $$I_2 = 8.13mA$$

Then the Voltage at B

$$I_2 = \frac{10-V_b}{675} = 8.13mA\\V_b = 4.51V$$

I just want to know if I am doing this question correctly. Any help would be appreciated it.

EDIT: After fixing my mistake, I found $$I_2 = 8.5 mA\\ V_b = 5.74V$$ And as others pointed out $$V_a = 5V$$

• What is R_eq, and how did you compute it? – Nick Johnson Oct 27 '15 at 7:38
• R_eq is the total equivalent resistor of all 4 resistors. Req = (675+500+1000)/(1175+1000) – Alp Oct 27 '15 at 7:39
• I don't think that's a useful figure here, because the current will not be evenly split between branches. Instead, consider each branch (R1+R3 and R2+R4) separately, and compute current and voltage for each. – Nick Johnson Oct 27 '15 at 7:42
• Yes I know. This is the total current which will split according to the ratio of the resistors. After splitting I get 10.37 mA on the left branch and 8.13mA on the right branch which are total of 18.5 mA – Alp Oct 27 '15 at 7:45
• Although I can't see a fault in your method I would also not go the Req way but consider the 2 branches separately since they're not influencing each other. Calculate power separately for each branch and use voltage divider formula to find voltage at b. At node a it's 5 V, I don't have to even calculate that. – Bimpelrekkie Oct 27 '15 at 7:47