Find the current I in this electrical circuit

[OmCheeto]

Would introducing Kirchhoff's voltage law make the problem too easy to solve?

 

[BvU]

View attachment 216507 For this matter, what everyone think, that BD is short circuit or not?

Not any more. Looks a lot differrent from the circuit in post #1 !

Here you simply draw the top right quarter circle as a horizontal line, erect the 6 $\Omega$ and see that it's parallel to the other 6 $\Omega$
V_DC= 6 V and Bob is the brother of your mother

 

[epenguin]

How do I manage to get this arithmetic wrong?! Yes we could just say the current through the 3 Ω resistor is $\frac{9.6}{3}$ = 3.2 A, and then the flow through the 9 Ω resistor is a third as much = (1 + 0.2/3) adding up to (4.2 + 0.2/3) = 4.26666... ≈ 4.27 A as you say.

 

[gneill]

If you want to work in fractions, write 9.6 as 96/10, then work through the fractions. Thus:

$I = \frac{96}{10} \left( \frac{1}{3} + \frac{1}{9} \right)\;A$

 

[aa_o]

If you want to work in fractions, write 9.6 as 96/10, then work through the fractions. Thus:

$I = \frac{96}{10} \left( \frac{1}{3} + \frac{1}{9} \right)\;A$

Just a note - working in fractions, or just exact for that matter, is very nice in mathematics, but when solving basic physics problems like these, i see no benefit - it can even obscure the result which with a reduced fraction givs 64/15 A. This does not give an intuitive knowledge of the 'normal' kind of values for these problems.

 

[epenguin]

If you want to work in fractions, write 9.6 as 96/10, then work through the fractions. Thus:

$I = \frac{96}{10} \left( \frac{1}{3} + \frac{1}{9} \right)\;A$

Yeah I thought of that too.

Just a note - working in fractions, or just exact for that matter, is very nice in mathematics, but when solving basic physics problems like these, i see no benefit - it can even obscure the result which with a reduced fraction givs 64/15 A. This does not give an intuitive knowledge of the 'normal' kind of values for these problems.

I used to think that too.

 

[gneill]

Just a note - working in fractions, or just exact for that matter, is very nice in mathematics, but when solving basic physics problems like these, i see no benefit - it can even obscure the result which with a reduced fraction givs 64/15 A. This does not give an intuitive knowledge of the 'normal' kind of values for these problems.

When you're comfortable with fractions, then I feel that the fraction result conveys the same information. Besides, you can always reduce the final result to a decimal version quickly enough.

 

[aa_o]

When you're comfortable with fractions, then I feel that the fraction result conveys the same information. Besides, you can always reduce the final result to a decimal version quickly enough.

Then I would highly recommend working symbolically, since this elucidates the relationships between the quantities in the configuration much clearer than does fractions. And this, too, will of course give the exact result if wished in the end.

 

[gneill]

Then I would highly recommend working symbolically, since this elucidates the relationships between the quantities in the configuration much clearer than does fractions. And this, too, will of course give the exact result if wished in the end.

Of course. Work symbolically as much as possible. In the end you'll still have an expression that needs to have values substituted and the math performed.

 

[Helly123]

If the 3 ohm and 9 ohm were series, then we found the answer 0,8 A which is on the options... so the 3 ohm and 9 ohm not series but parallel?

 

[BvU]

Of which diagram ?

 

[Helly123]

Of which diagram ?

The #1 diagram

 

[aa_o]

You can see that they are en parallel, since both of the resistor terminals (of the 3 and 9 ohm resistors) share the same nodes.
If there's no option for giving around 4.27 A as an answer, there is a mistake in the options.

 

[BvU]

This one

equals

equals

Which for some unspecified reason still shows a resistor between B and C in the post but not in the picture attached (below riight)

Obviously parallel , right ?

 

[Asymptotic]

Yet another way to think about it.

If this handful of resistors and 9.6V battery were wired together it would be as though a jumper wire was connected across the 6Ω resistor.

What are the current path(s)?

A useful formula valid when two (and only two!) resistors are connected in parallel is R_t=(R_a*R_b)/(R_a+R_b).

 

[Helly123]

Thanks guys. How about this?

Is the 6o hm (A) short circuit (jumper wire)?

Is the BD short circuit (jumper wire)?

 

[aa_o]

Thanks guys. How about this? View attachment 216636
Is the BD short circuit (jumper wire)?

Can you draw a line from B to D without going through any components?

 

[Helly123]

Can you draw a line from B to D without going through any components?

I suppose, i can not, it goes through AB. Isnt it?

 

[aa_o]

I suppose, i can not, it goes through AB. Isnt it?

Yes, so it is not shorted.

 

[Helly123]

Yes, so it is not shorted.

View attachment 216636

Is the 6o hm (A) short circuit (jumper wire)?

 

[Asymptotic]

View attachment 216636

Is the 6o hm (A) short circuit (jumper wire)?

Let's ask this in another way. Is the voltage the same at both sides of the resistor?

Start at the positive battery terminal, and trace the circuit with your finger ...
Label the 6 ohm resistor "R_A", one side of the resistor Z, and the other side Z'.
What is the voltage at Z?
9.6 volts.

And at Z'?
Also, 9.6 volts.

What does Ohm's law tell us? If the voltage is the same at both sides of the resistor (that is, voltage drop across the resistor is zero) then IR_A=VR_A/R_A = 0 volts/6 ohms = 0 amps.

In effect, 6 ohm resistor R_A is shorted by the "wire" between point Z and Z'.

In this drawing it's the same for connections between R_A and R_B. The only component with a voltage drop across it is the 9 ohm resistor R_C.

Let's re-state the problem in yet another way.
R_A is a 6 ohm resistor in parallel with a 0 ohm wire between point Z and Z'.
The general rule is that total resistance for two resistors in parallel is less than the smaller resistor. In this special case the smaller resistor is 0 ohms, and limits total resistance to 0 ohms.

 

[Helly123]

Thank you for the explanation :)

 

[Helly123]

In this drawing it's the same for connections between R_A and R_B. The only component with a voltage drop across it is the 9 ohm resistor R_C.

So both A and B are short circuit, because the voltage stream in left side and right side are zero. I see.