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 $$V_a = 5V$$

  • \$\begingroup\$ What is R_eq, and how did you compute it? \$\endgroup\$ – Nick Johnson Oct 27 '15 at 7:38
  • \$\begingroup\$ R_eq is the total equivalent resistor of all 4 resistors. Req = (675+500+1000)/(1175+1000) \$\endgroup\$ – Alp Oct 27 '15 at 7:39
  • \$\begingroup\$ 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. \$\endgroup\$ – Nick Johnson Oct 27 '15 at 7:42
  • \$\begingroup\$ 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 \$\endgroup\$ – Alp Oct 27 '15 at 7:45
  • \$\begingroup\$ 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. \$\endgroup\$ – Bimpelrekkie Oct 27 '15 at 7:47

You're overcomplicating things by finding the equivalent resistnace for the whole circuit, then having to determine what proportion of current flows down each branch. Instead, compute total resistance for the branch R2+R4, determine the current down that branch, and from that determine the voltage across R4 to get the voltage at point b. For the other branch, you can simply observe that both resistors are the same, so the voltage will be exactly half the supply voltage.

Other than that, and what looks like a math error, your method seems sound.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.