-2
\$\begingroup\$

I would like to know why Ohm's law and Kirchhoff laws give different results for voltage as said in this lecture(p.19) by Dr. Sheryl Howard.

Even though the change is small in voltage it gave me very different results for current.These are the results obtained using Ohm's and Kirchoff law:

\$10A\centerdot R_{eq} = 8I_1 + 4I_2 - (1)\$ (Using Ohm's law)

\$8I_1 - 4I_2 = 0 - (2)\$(Using KVL)

Solving these we get \$I_1 = 1.67A\$ while the value for \$I_1\$ in the lecture is \$3.33A\$.

Could anyone tell why this change happened for current and voltage.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – W5VO Apr 11 '16 at 16:17
4
\$\begingroup\$

I don't know how you came to the conclusion that Ohm's Law and Kirchhoff's Laws give different solutions (maybe it does (which I doubt), but really not in this case); the lecture's answer is right.

Proof:

The total resistance of two parallel resistors is:

$$R_t = \frac{R_1*R_2}{R_1 + R_2}$$

therefore: $$\frac{8\Omega*4\Omega}{8\Omega + 4\Omega} = 2.667 \Omega$$

Current in parallel circuits is reciprocal to the proportions of the resistors:

$$\frac{R_1}{R_2} = \frac{I_2}{I_1}$$

which gives $$\frac{8}{4} = \frac{I_2}{I_1}$$

and $$I_1=\frac{I_2}{2}$$

on the other side, the lecture states that the whole current is 10A, so

$$I_1 + I_2 = 10\text{A}$$

which gives $$\frac{I_2}{2} + I_2 = 10\text{A}$$ $$I_2 + 2*I_2=2*10\text{A}$$ $$I_2(1+2) = 20\text{A}$$ $$I_2 = \frac{20\text{A}}{3} = 6.667\text{A}$$

and $$I_1 = 3.333\text{A}$$

Now, we have all resistor values and the current through each resistor, we can continue with voltage (and Ohm's law):

$$V_1 = R_1 * I_1$$ $$V_1 = 8\Omega * 3.333\text{A} = 26.667\text{V}$$ $$V_2 = R_2 * I_2 = 4 * 6.667 = 26.667\text{V}$$ (this has to be because in a parallel circuits, voltage is the same on all parallel components, current is different)

Now we can consider the source. Because voltage is the same on every parallel component, the source gives 26.667V. It delivers 10A, therefore

$$R_{eq}=\frac{V_{eq}}{I_{eq}}=\frac{26.667\text{V}}{10\text{A}} = 2.667\Omega$$

Simple calculation shows (with no complex components) that the solution in the lecture (I1 = 3.333A) was right and that Kirchhoff's laws (10A = 6.667A + 3.333A) (at least in this example) don't conflict with Ohm's law.

regards

\$\endgroup\$
  • \$\begingroup\$ +1. Good answer. I cleaned up your equations to make them consistent with the conventions that are typically found on EE.SE: "." as a decimal separator and "V" for voltage. Also, some of your units were muddled, hopefully I didn't introduce any errors. \$\endgroup\$ – uint128_t Apr 11 '16 at 15:48
  • \$\begingroup\$ I still had the german measuring units in mind, so it seems correct now. Thank you. \$\endgroup\$ – Coliban Apr 12 '16 at 4:19
  • \$\begingroup\$ I don´t know, why the OP was downgraded, four times. This question is obviously a question which comes up, when students begin a new theme in electornics and, although, the OP made a small mistake, the question is a good example for anybody willing to get into the electronic realm. I think, people should not be discouraged when they try to understand a non trivial problem by downgrading them. \$\endgroup\$ – Coliban Apr 14 '16 at 6:04

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.