Voltage query -- A power supply, resistors and a voltmeter....

[alsy]

Homework Statement

a 12v dc supply is connected across 2 100k resistors in series.a voltmeter having sensitivity of 10,000 ohms per volt is switched to its 10v range and is connected to its 10v range and is connected to measure the voltage of 1 resistor.

Homework Equations

1)sketch circuit
2)calculate voltage reading
3)calculate percentage error of the meter reading in 2

The Attempt at a Solution

I can draw the circuit that part is fine.im using v=ir to try and workout voltage so 100,000/10000 = 10
and for the percentage error 100/10 =10%

 

[gneill]

Homework Statement

a 12v dc supply is connected across 2 100k resistors in series.a voltmeter having sensitivity of 10,000 ohms per volt is switched to its 10v range and is connected to its 10v range and is connected to measure the voltage of 1 resistor.

Homework Equations

1)sketch circuit
2)calculate voltage reading
3)calculate percentage error of the meter reading in 2

Um, those aren't relevant equations. That's just more of the problem statement.

For relevant equations you want to state things like Ohm's law, or the formulas for combining resistors, or the voltage divider equation, and so forth. Whatever you think might be applicable to this type of problem.

The Attempt at a Solution

I can draw the circuit that part is fine.im using v=ir to try and workout voltage so 100,000/10000 = 10
and for the percentage error 100/10 =10%

Can you show your work in detail? How did you determine the resistance of the meter? How did you come up with "100,000/10000 = 10", and what does it mean? What are the units?

 

[alsy]

sure thing,i would be using ohms law.100,000r ohms /10,000 I =10volts reading

for the percentage error I would say 100r / 10v =10 % error

 

[gneill]

sure thing,i would be using ohms law.100,000r ohms /10,000 I =10volts reading

for the percentage error I would say 100r / 10v =10 % error

Sorry, I'm not understanding where the values are coming from, and I'm not certain what the r, v, and I are meant to mean.

Can you draw a circuit diagram and label the parts with their values?

 

[alsy]

attach

 

[gneill]

Okay, what resistance value are you going to assign to the meter (Let's call it Rm)? Remember, it's rated at 10,000 Ohms per volt and it's on the 10 Volt scale...

 

[alsy]

ok rm I would say 10,000 x 12dc v = 120,000/10v =12,000r

 

[gneill]

ok rm I would say 10,000 x 12dc v = 120,000/10v =12,000r

Not quite. It's on the 10 V scale. The source voltage V1 (12 V) doesn't play a part in the meter's internals. The meter's resistance will depend upon its Ohms-per-Volt rating and the scale selected. So

##R_m = \frac{10 kΩ}{V} 10 V = 100 kΩ##

That is the resistance that the meter will place across R2 when it takes a reading there.

So now, what calculations will you perform to find the voltage that the meter reads?

 

[alsy]

I would work out the volt across the 1 resistor battery voltage/total resistance x r1 = voltage

 

[gneill]

I would work out the volt across the 1 resistor battery voltage/total resistance x r1 = voltage

Okay, go ahead. Show us the details.

What will you do with that voltage across R1?

 

[alsy]

12v /200k x 100k =6volts across r1

I would be reading that with the voltmeter as my voltage reading

 

[gneill]

How did you arrive at 200k for the total resistance?

 

[alsy]

100 +100 =200k resistors in series rt=r1+r2

 

[gneill]

100 +100 =200k resistors in series rt=r1+r2

Nope. You've left out the meter resistance. It's in parallel with R2.

 

[alsy]

yes your absolutely right. 100k + 1/100 +1/100=150k ohms

 

[gneill]

yes your absolutely right. 100k + 1/100 +1/100=150k ohms

I understand what you're trying to convey, but your math exposition is a tad sloppy. If I ran that expression through my calculator as it is written then I'd arrive at 100.02 k Ohms, not the 150 k Ohms that you intend. You can use parentheses to gather terms, and the edit panel provides icons for doing things like superscripts and subscripts:

100k + (1/100 +1/100)^-1 = 150k ohms

Now, 150 k Ohms for the total resistance is good. What resistance do you want to know the potential across? You could find the potential across R1, or take the more direct route and find the potential difference across the R2||Rm parallel pair. After all, you've calculated the resistance of that pair, right?

 

[alsy]

well the question is asking for the voltage reading across r1

so 12v / 150k ohms x 100k ohm =0.008v so this would be the reading

 

[gneill]

well the question is asking for the voltage reading across r1

Well, no, it's asking for the voltage reading on the meter. The meter is connected across R2 (according to your figure):

so 12v / 150k ohms x 100k ohm =0.008v so this would be the reading

Use parentheses to gather your terms to force the correct order of operations:

(12v / 150k ohms) x 100k ohm = ?

That would be the voltage across R1. Of course, you want the voltage across the R2||Rm pair. You could recalculate that or use KVL to figure it out from V1 and this VR1.

 

[alsy]

(12v / 150 ohms) x 100k ohm =8volts

kvl??

(12v / 100k) ohms x 100k ohms = 12volts

 

[gneill]

(12v / 150k ohms) x 100k ohm =8volts

Right. That's the voltage drop across R1.

kvl??

Kirchhoff's Voltage Law. You should look that up. It's very important in circuit analysis! You'll use it very often indeed, along with KCL, Kirchhoff's Current Law.

(12v / 100k) ohms x 100k ohms = 12volts

I don't know what that calculation is... The 100k in the denominator is not the total resistance, and the 100k looks like R1 again...

It would be preferable if you would first present the calculation as symbols rather than numbers. That way we can tell what it is the calculation represents.

 

[alsy]

12v / 150k ohms) x 100k ohm =8volts is the reading across R2 with the voltmeter

Kirchhoff's,thanks I will look this up.

 

[gneill]

12v / 150k ohms) x 100k ohm =8volts is the reading across R2 with the voltmeter

No, the voltmeter's 100k is in parallel with R2's 100k. So the current splits between the two resistances, and the resulting voltage drop across the parallel combination will be less. What's the net resistance of Rm||R2 (that's Rm in parallel with R2)?

 

[alsy]

(1/100k + 1/100) = 50,000 ohms resistance

 

[gneill]

(1/100k + 1/100) = 50,000 ohms resistance

You mean:

1/(1/100k + 1/100) = 50,000 ohms resistance

Don't forget the overall reciprocal.

Resistors in parallel:
$$R_p = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n} } $$
or, stated another way,
$$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n} $$

So, what do you calculate for the voltage that the meter will read?

 

[alsy]

1/(1/100k + 1/100) = 100,000ohms resistance

 

[gneill]

1/(1/100k + 1/100) = 100,000ohms resistance

You'll need to explain what you're doing. Random (incorrect) calculations with numbers aren't very helpful.

 

[alsy]

1/
(resistor 1 in parallel )1/100k + (voltmeter in parallel ) 1/100) = 100,000ohms resistance with voltmeter connected in r2

 

[gneill]

1/
(resistor 1 in parallel )1/100k + (voltmeter in parallel ) 1/100) = 100,000ohms resistance with voltmeter connected in r2

Can you explain in words what you're trying to accomplish with that calculation? I don't see resistor R1 being in parallel with anything, so "(resistor 1 in parallel )" confuses me. When I look at the circuit I see resistor R1 in series with the parallel combination of R2 and Rm. Look at the equivalent circuit that I posted above (post #22).

From now on, don't just show me numbers. Show me symbols first. Use R1, R2, Rm and V1. Add variables as required, but define them. So you might say that for the parallel combination of R2 with the meter's resistance Rm that Rp = R1||Rm, which is Rp = 1/( 1/R1 + 1/Rm) = 50k Ohms. Then you'd have the variable Rp to use for that value.

 

[alsy]

1/
(R2 in parallel )1/100k + (RM ) 1/100) = 50,000ohms resistance with voltmeter connected in r2

So (r1)100k ohms + (R2|Rm) 50k ohms =150k ohms.

 

[gneill]

1/
(R2 in parallel )1/100k + (RM ) 1/100) = 50,000ohms resistance with voltmeter connected in r2

So (r1)100k ohms + (R2|Rm) 50k ohms =150k ohms.

Write:

Parallel resistance of R2 with the meter resistance Rm:
##Rp = R1 || Rm##
##Rp = 1/ (1/100k + 1/100k) = 50k## Ohms

Total circuit resistance with meter in place:
##Rt = R1 + Rp##
##Rt = 100k + 50k = 150k## Ohms

That should help make things clear.

Now, can you show a calculation for the voltage across the part of the circuit where the meter is connected?

 

[alsy]

ok battery voltage/total resistance x R2= voltage across r2

12v / 150k ohms x 50k ohms = 4 V

 

[gneill]

ok battery voltage/total resistance x R2= voltage across r2

12v / 150k ohms x 50k ohms = 4 V

Yes! Well done.

So the meter will read 4 V. That answers part 2 of the question. Now, how will you address part 3?

 

[alsy]

thank you.

ok part 3.

10,000ohms sensitivity x 4v= 40,000 /10v range = 4000 / 100= 40%

 

[gneill]

I think what they're looking for is the percent error in the measured value when compared to what an ideal meter would read (that is a meter which would not affect the circuit operation at all, one with infinite ohms per volt sensitivity).

What would the voltage be across R2 if the meter was not attached?

 

[alsy]

12v / 200k ohms x 100k ohms = 6v voltage across r2 without meter