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[JS_Electric]

Yesterday, I faced an interview. The interviewer asked the question....

Que. The parallel combination of a 470 resistor and a 1.5 k resistor is in series with the parallel combination of five 1 k resistors. The source voltage is 50 V. What will be the percentage of the load current through any single 1 k resistor?

 

[Des 56]

You have equal value resistances sharing 100% of the current

 

[topquark]

It's 20%.

 

[Geoff]

20%

 

[infinity]

is this because the 1K is least resistive path meaning the current will split across the 5 1K resistors?

 

[topquark]

The first bit, the 470 and 1.5K in parallel you can ignore, as the total current will be the same in and out of that bit of the circuit. So the load current is then evenly split between the 5 in parallel resistances. As all 5 are all the same resistance value; the current will flow evenly between them.

 

[Des 56]

It's 20%.

I didn't give a figure because I understand what will happen in practice,but its the maths that's defeating me :38:

 

[topquark]

Two parallel sets of components the 470 + 1.5K Ohm (as only two you could just work out as product/sum). As it's in series with the 5 parallel resistors you don't really need to work that bit out as the total current will now be available again to drive the 5 parallel resistors.

So that's the old recip sum = 1/(1/1 + 1/1 + 1/1 + 1/1 + 1/1) = 1/5 (0.2) Ohms total. That through each resitor is therefore 1/5th of the total ie 20% (a little harder but same method if the resistors are significantly different in value).

 

[Silly Sausage]

or generally:

R[SUB]T[/SUB] = 1/(1/R1 + 1/R2 + ... + 1/Rn)

I[SUB]Rx[/SUB] = I[SUB]total[/SUB] * (R[SUB]T[/SUB] / Rx)

 

[Knobhead]

If you want to be pedantic it’s 0.0179A through each leg of the 5. But that wasn’t the question.

 

[Des 56]

I didn't give a figure because I understand what will happen in practice,but its the maths that's defeating me :38:

Thank you topquark
I feel guilty now, I was really only joking about the maths :26:

 

[topquark]

Thank you topquark
I feel guilty now, I was really only joking about the maths :26:

I had a sneaking suspicion you were; given some of our previous conversations You never know, it might confuse someone one day

 

[malcolmsanford]

Reading this thread I really wished I had paid more attention when at college they gave us the 35 minute training course on electronics .........but at the time I was too busy training to be a rag a**e to take much notice of these piddling pesky little things...................... top answers there lads

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Reading this thread I really wished I had paid more attention when at college they gave us the 35 minute training course on electronics .........but at the time I was too busy training to be a rag a**e to take much notice of these piddling pesky little things...................... top answers there lads

 

[i=p/u]

current is the same in parallel, so that rules that out, only leaving you series which splits according to size of slice. this is a good way saying i understand and was good at this stuff one time. but obv not as bright as yous

 

[ringer]

current is the same in parallel, so that rules that out, only leaving you series which splits according to size of slice. this is a good way saying i understand and was good at this stuff one time. but obv not as bright as yous

?????

Current is not the same in parallel. Volt drop is the same, and current varies according to the resistances. If all resistances in parallel are the same, then the current is split evenly in each branch.