how is this worked out? ohms law question

[gjpollitt]

Anyone know how to work out the current in the circuit? I know the diagram gives the answer of 2.5ma but how do you calculate that when you dont know the value of R3?

I thought you would need 2 out of 3 variables to calculate the third ? ie know the R and I to calculate V drop across Rx.

 

[stevez]

I'll give you a bit of a clue - you know that 20 volts is applied and that 4.5 volts is dropped across R3 - leaving 15.5 volts total drop across R1 and R2. You can determine the voltage drop across R1 or R2 because you have the values of R - just do the math.

Was that of any help?

 

[checkmate]

You must be very new to electronics to ask such a question, so i'll try to explain it as simple as possible.

A voltage (or potential) drop is always calculated with respect to a start and end point. The circuit is a loop, so the start and end points are the same, hence net voltage drop around the loop is also zero.

If the battery gives voltage rise of 20V, and R3 give voltage drop of 4.5V, then R1 & R2 combined should give a voltage drop of 15.5V. The rest is ohm's law.

This is a simple circuit. There are times whereby the circuit is more complicated, and you would have to use variables to replace unknowns and solve via simultaneous equations.

Here's a good link for multiple loop systems.
**broken link removed**

 

[gjpollitt]

If the battery gives voltage rise of 20V, and R3 give voltage drop of 4.5V, then R1 & R2 combined should give a voltage drop of 15.5V. The rest is ohm's law.

Ok thanks for replying.

If R=VI and you need to solve for R3 knowing R1 and R2 and also that the supply voltage is 20v how do you do it? You have v=20 but where do you find the I? I could work out for R1 and R2 as I know the values of those but if you add in R3 as an unknown surely you cant work out the total current?

sorry if this a silly question but I cant get my head around it

thanks

 

[checkmate]

Firstly, you got Ohm's Law wrong. It's V=RI.
Kirchoff's voltage law states that sum of all voltages in a loop must sum to zero.
We just sum up all voltages, starting from the battery. Since battery gives voltage (instead of dropping it), we give it a negative voltage drop, ie v(batt)=-20V

Now, V(Batt)+V(R1)+V(R2)+V(R3)=0
The unknowns are V(R1) and V(R2), so you use Ohm's Law on them.
V(Batt)+R1I+R2I+V(R3)=0.
So just solve for I.

 

[gjpollitt]

ok I understand ohms law etc but the original question on that subject (i have an interactive tutorial on electronics which the above diagram was taken from) asked myself to calculate value of R3. The current was not on the diagram at that time as the values are displayed as I answer the questions so ignore the 2.5ma for now.

Therefore going from what you just told me, again, how do you calculate a value of resistor R3 when all you know is the voltage drop across it but not the current.

Using ohms law (I=V/R) what is R? (R=V/I) but what is I? Do you see what I mean?

I can work out the current through R1 and R2 using just values of Rx and V but not R3.


Thanks again

edit - in fact I can see the confusion from my original question. Given that current is 2.5ma it is easy to calcuate Vx across any Rx but the 2.5ma was not given to me when the original question was asked on the test so I need to know if the test is incorrect or if there is some other way of calculating 2 unknowns when only given 1 value.

 

[eblc1388]

You don't need to know the value of R3 to work out the current.

R1 and R2 see a net voltage of (20-4.5)=15.5V, knowing their resistance values, you can work out the current immediately, using Ohm's Law.

Because now you know the current, a resistor dropping 4.5V at that current is then a piece of cake to find out its resistance value, again using Ohm's Law.

Got it? By the way, the conventional current direction is wrong in your first diagram.

 

[gjpollitt]

got it, the value is 1.8k

20-4.5=15.5

i=v/r =15.5/6200 = 2.5ma

r=v/i =4.5/2.5ma=1.8k

thanks for your help all

 

[eblc1388]

got it, the value is 1.8k

thanks for your help all

Excellent.

 

[gjpollitt]

another question

How would I solve this for R2?

I cant find the voltage drop across R1 as I cant use the formula

VR1=(R1/Rt)*Vt

as I dont know what Rt is so how would I also find out what R3 is to calculate R2?

thanks again

 

[eblc1388]

You ask yourself the question:

If I have a 19V battery and I want a single resistor to pass a current of 2mA, which value would I use?

Then you search your component box and find out you have two resistors of values 4.7K and 1.5K which you can join together but then you cannot make up the value you wanted. You need another resistor, so what would be the value of this resistor?

 

[gjpollitt]

thanks that it is a good way of looking at it

so R3 would be r=v/i = 19/.002 = 9500 - r1 -r3 = 3.3k