need help please why my textbook says that there is no current across both resistor

[07kim36]

can anybody explain this to me.. why theres no current flow across R1 and R2?

 

[ericgibbs]

can anybody explain this to me.. why theres no current flow across R1 and R2?

Hi,.
Do you know how to calculate the sum/result of resistors in parallel.??

The short circuit is considered as zero ohms.

E,

 

[DerStrom8]

You can find parallel resistance by the formula,

1/(1/R1 + 1/R2 + 1/R3)

Use the resistances given for R1 and R2 in the formula, and use 0 ohms for R3. Notice something? You're trying to divide 1/0. That effectively gives you infinity ohms. What's happening is that all the current is going through the short (since it's the path of least resistance--in theory no resistance at all--so the current isn't going to go through elements with a higher resistance). Make sense?

Matt

 

[07kim36]

Hi,.
Do you know how to calculate the sum/result of resistors in parallel.??

The short circuit is considered as zero ohms.

E,

everyone knows how to calculate resistor ... but my question is why theres no current flow in I1 and I2..??? why I1 =0A ?? why I2=0A???? this topic is on the boylestad book concerning short and open circuit...

 

[07kim36]

You can find parallel resistance by the formula,

1/(1/R1 + 1/R2 + 1/R3)

Use the resistances given for R1 and R2 in the formula, and use 0 ohms for R3. Notice something? You're trying to divide 1/0. That effectively gives you infinity ohms. What's happening is that all the current is going through the short (since it's the path of least resistance--in theory no resistance at all--so the current isn't going to go through elements with a higher resistance. Make sense?

Matt

awww... haha.. ^^ TQ .. now i get it... omg... thanks bro

 

[07kim36]

sir .. does that also means that theres no voltage drop across the resistor r1 and r2?

 

[ericgibbs]

You can find parallel resistance by the formula,

1/(1/R1 + 1/R2 + 1/R3)

Use the resistances given for R1 and R2 in the formula, and use 0 ohms for R3. Notice something? You're trying to divide 1/0. That effectively gives you infinity ohms. What's happening is that all the current is going through the short (since it's the path of least resistance--in theory no resistance at all--so the current isn't going to go through elements with a higher resistance. Make sense?

Matt

hi,
I thought we agreed that we would not give direct answers to Homework questions, only hints or clues.??

E.

 

[07kim36]

hi,
I thought we agreed that we would not give direct answers to Homework questions, only hints or clues.??

E.

i opologies for my rude behaviour... please forgive me...
and for the next question .. is there a voltage drop across those resistor?

 

[07kim36]

i am guessing theres 0v voltge drop across both of em' ..am i right?

 

[ericgibbs]

i opologies for my rude behaviour... please forgive me...
and for the next question .. is there a voltage drop across those resistor?

hi,
No problem.

If you had calculated the total resistance including the short circuit of Zero Ohms, then used Ohms Law it would have shown you that you have Zero volts across the resistor network, hence no current.

A point to study would be the effect of the voltage driving sources internal output resistance.

All voltage sources have some internal resistance, even though it maybe very low, this would limit the short circuit current.

E.

 

[DerStrom8]

hi,
I thought we agreed that we would not give direct answers to Homework questions, only hints or clues.??

E.

Sorry about that Eric. I started off by writing a clue, but I guess it kind of turned into the answer....

I think the OP learned from it though, don't you?

i am guessing theres 0v voltge drop across both of em' ..am i right?

For that use Ohm's Law: V=I*R.

Matt

 

[07kim36]

hi,
No problem.

If you had calculated the total resistance including the short circuit of Zero Ohms, then used Ohms Law it would have shown you that you have Zero volts across the resistor network, hence no current.

A point to study would be the effect of the voltage driving sources internal output resistance.

All voltage sources have some internal resistance, even though it maybe very low, this would limit the short circuit current.

E.

thank you sir...and i am guessing that theres an infinite resistance in a voltmeter right? because if theres no resistance in the voltmeter then i assume it will be like a short circuit .

 

[ericgibbs]

thank you sir...and i am guessing that theres an infinite resistance in a voltmeter right? because if theres no resistance in the voltmeter then i assume it will be like a short circuit .

hi,
Modern volt meters have a very high input impedance/resistance, so that when used to measure a circuit voltage they draw a VERY small current from the circuit, so they do not load the circuit.

You should consider all 'short circuits' have some resistance even though it can be a very low resistance/

Your voltage source on the problem circuit would have a high resistance compared to the 'short circuit', otherwise it would have supplied enough current thru the short circuit to melt the wire.

Is this what you mean.?

E

 

[MrAl]

You can find parallel resistance by the formula,

1/(1/R1 + 1/R2 + 1/R3)

Use the resistances given for R1 and R2 in the formula, and use 0 ohms for R3. Notice something? You're trying to divide 1/0. That effectively gives you infinity ohms. What's happening is that all the current is going through the short (since it's the path of least resistance--in theory no resistance at all--so the current isn't going to go through elements with a higher resistance. Make sense?

Matt

Hi there Matt,

I like the way you approached this question. And if you take your three level formula and reduce it to just two levels you get:

Rparallel=(R1*R2*R3)/(R2*R3+R1*R3+R1*R2)

and from this it is immediately clear that Rparallel must be equal to zero when R3 is made equal to zero and we dont run into any mathematical anomalies

Rparallel=(R1*R2*R3)/(R2*R3+R1*R3+R1*R2)
Rparallel=(R1*R2*R3)/(R2*0+R1*0+R1*R2)
Rparallel=(R1*R2*R3)/(R1*R2)
Rparallel=(R1*R2*0)/(R1*R2)
Rparallel=0/(R1*R2)
Rparallel=0

 

[JoeJester]

07kim,

consider the below example.

The Ri is the internal resistance of the battery. You have two resistors in parallel and the wire with a very small resistance shorting out the two parallel resistors. If you hooked up your meter to measure the voltage acrossed the resistor you would read a very small value. If you hooked up an ammeter to measure the current through the resistor, you would read a very small value. If you decreased the value of the internal resistance and the resistance of the wire shorting across the two parallel resistors, well, you can configure an excel spreadsheet to calculate them out as easily as I can so you can instantly see how the resistances affect the circuit.

A minor reminder that voltage is measured acrossed a resistor and current flows through a resistor.

 

[DerStrom8]

07kim,

consider the below example.

The Ri is the internal resistance of the battery. You have two resistors in parallel and the wire with a very small resistance shorting out the two parallel resistors. If you hooked up your meter to measure the voltage acrossed the resistor you would read a very small value. If you hooked up an ammeter to measure the current through the resistor, you would read a very small value. If you decreased the value of the internal resistance and the resistance of the wire shorting across the two parallel resistors, well, you can configure an excel spreadsheet to calculate them out as easily as I can so you can instantly see how the resistances affect the circuit.

A minor reminder that voltage is measured acrossed a resistor and current flows through a resistor.

Hey Joe, good to see you over here! (I assume you're the JoeJester from AAC....?)

 

[JoeJester]

Yes, the same as AAC. I just hadn't updated my location till just now as Michigan was my former location.