How to calculate the total resistance in this circuit?

[Adel Makram]

Homework Statement

How to calculate the value of total resistance R_T in the shown circuit?

Homework Equations

The Attempt at a Solution

I used Kirchhoff` first law: i<sub>1</sub>+i<sub>2</sub>=I where i<sub>1</sub>, i<sub>2</sub> and I are the current passing through R<sub>1</sub>, R<sub>2</sub> and the total current in the circuit, respectively.
similarly, i<sub>4</sub>+i<sub>5</sub>=I,
i<sub>1</sub>+i<sub>3</sub>=i<sub>4</sub>
i<sub>3</sub>+i<sub>5</sub>=i<sub>2</sub>
Then I used Kirchhoff` second law,
20 i₁+10 i₄=V₀
10 i₂+20 i₅=V₀.

 

[phinds]

Given that you have not specified any points from which to take the Rth, you have an incomplete problem statement and as such it has no answer.

 

[Adel Makram]

Given that you have not specified any points from which to take the Rth, you have an incomplete problem statement and as such it has no answer.

I edited the figure after puting labels of currents and points of intersection.

 

[Adel Makram]

I tried this as well,
considering the contour that includes i₁ and i₄
$$ 20 i_1 + 10 i_4=V_0 $$
considering the contour that includes i₂ and i₅
$$ 10 i_2 + 20 i_5=V_0$$
multiplying the second equation by 2 and collect term $$(i_1 + i_2)$$.
$$ 20(i_1 + i_2) + 10 i_4 +i_5=3V_0 $$
divide on $$ I=i_1+i_2 $$ and consider $$ R_T=\frac{V_0}{I} $$
$$ 3R_T=20+\frac{10 i_4}{I}+\frac{40 i_5}{I} $$

 

[phinds]

I edited the figure after puting labels of currents and points of intersection.

And you STILL haven't specified any points on which to base an Rth.

 

[Adel Makram]

And you STILL haven't specified any points on which to base an Rth.

What do you mean by specific point to base an Rth?

 

[phinds]

What do you mean by specific point to base an Rth?

Do you understand what a Thevenin Equivalent circuit IS? When your question said find Rt, and there is no Rt specified in your diagram, I assumed you meant Rth, the Thevenin Equivalent resistance. Did you mean something else?

 

[Adel Makram]

I meant the total resistance of the circuit.

 

[phinds]

I meant the total resistance of the circuit.

Based on what ?

If you mean what is seen by the power supply, you have to say so. If that is what you mean then you would profit by looking up delta-Y transforms.

 

[Adel Makram]

That is great thank you.
So transforming Δ circuit to Y circuit as in the diagram,
Transforming the Δ component on the right side of the circuit to Y-component yields,
$$R_1=\frac{R_a R_c}{R_a+R_b+R_c}$$
$$R_2=\frac{R_b R_c}{R_a+R_b+R_c}$$
$$R_3=\frac{R_a R_b}{R_a+R_b+R_c}$$
so, $$R_1=\frac{10}{4}$$
$$R_2=5$$
$$R_3=5$$
simplifying the circuit now yields,
the left side of the circuit is a parallel circuit while the right side is series one.
the resistance of the left side is $$R=\frac{(20+\frac{10}{4})(15)}{(20+\frac{10}{4})+(15)}=9$$
summing to the right side of the circuit with a R=5 yields 9+5=14.

 

[phinds]

I didn't check your calculations, but that's definitely the right way to do it.