Physics Electricity Potential Dividers & Resistance [PLEASE HELP]

Hi,
This is a question I found in the textbook [see attached]
Now, here are my current solutions so far.

Part A
i) 1.0KΩ + 1.0KΩ = 1000Ω + 1000Ω = 2000Ω [For later calculation, the answer is kept as Ω]
ii) 1000Ω + Effective Parallel Resistance.

\frac{1}{R_T} = \frac{1}{1000} + \frac{1}{R}

This simplifies to

\frac{1}{R_T} = \frac{R}{1000R} + \frac{1000}{1000R} = \frac{1000 + R}{1000R}

so R_T (parallel) =

\frac{1000R}{1000 + R}

So the final resistance for part ii is

1000 +\frac{1000R}{1000 + R}

Part B
There are two ways. First I will find the current because this information becomes necessary in part C.

I = \frac{V}{R} = \frac{6}{2000} = 3 \times 10^{-3}A\

Then

Unparseable latex formula:

V \ = \ IR = 3 \times 10^{-3}A \times 1000 \ohm = 3V

ALTERNATIVE

V_{out} = \frac{1000}{1000 + 1000} \times 6 = 3V[b] [/b]

Part C
I already calculated that the current is

3\times 10^{-3}A

Using

\frac{V}{I} = R = \frac{2}{3\times 10^{-3}A} = 666.7 \Omega

This means that

\frac{1}{666.7} = \frac{1}{1000} + \frac{1}{R}

or

\frac{1}{666.7} = \frac{1000 + R}{1000R}

How can I finish this question to find R?
Please help.

Answers are now included for reference.

(edited 8 years ago)

Screen Shot 2016-04-08 at 23.17.04.png396.8KB

Reply 1

username1560589

20

I haven't read through your working, but assuming your last line is correct then it's just a linear equation in R.

Reply 2

nwmyname

OP

13

Original post by morgan8002

I haven't read through your working, but assuming your last line is correct then it's just a linear equation in R.

I tried that, and I didn't get anywhere close to the correct answer. I got a very small negative value.

Reply 3

nwmyname

OP

13

@morgan8002 Answers are here.

Screen Shot 2016-04-09 at 11.08.49.png20.1KB

Reply 4

Nadile

13

Original post by nwmyname

I tried that, and I didn't get anywhere close to the correct answer. I got a very small negative value.

Original post by nwmyname

@morgan8002 Answers are here.

You must have messed up your maths there to get a very small negative because you should get the correct answer from your working out.

Reply 5

uberteknik

21

Original post by nwmyname

Hi,
This is a question I found in the textbook [see attached]
Now, here are my current solutions so far.

Part C
I already calculated that the current is

3\times 10^{-3}A

Using

\frac{V}{I} = R = \frac{2}{3\times 10^{-3}A} = 666.7 \Omega

This means that

\frac{1}{666.7} = \frac{1}{1000} + \frac{1}{R}

or

\frac{1}{666.7} = \frac{1000 + R}{1000R}

How can I finish this question to find R?
Please help.

Answers are now included for reference.

You have made an incorrect assumption.

With the voltmeter added to the circuit, the total current can no longer be 3mA because the parallel resistance with the voltmeter added must now be less than 1K ohm.

R_total = 1000 + R_parallel

Since R_parallel is now less than 1K

R_total = 1000 + (<1000) = <2000 ohms

i.e. the current leaving and entering the supply must now be greater than 3mA.

HINT: The p.d. developed across the top series resistor is 4V (6 - 2) with the voltmeter added to the circuit.

(edited 8 years ago)

Reply 6

nwmyname

OP

13

Original post by uberteknik

You have made an incorrect assumption.

With the voltmeter added to the circuit, the total current can no longer be 3mA because the parallel resistance with the voltmeter added must now be less than 1K ohm.

R_total = 1000 + R_parallel

Since R_parallel is now less than 1K

R_total = 1000 + (<1000) = <2000 ohms

i.e. the current leaving and entering the supply must now be greater than 3mA.

HINT: The p.d. developed across the top series resistor is 4V (6 - 2) with the voltmeter added to the circuit.

Please check the solutions as this is not stated.

Screen Shot 2016-04-09 at 11.08.49.png20.1KB

Reply 7

JN17

10

It looks like you get the correct answer with the assumption that the current is the same, even though it isn't, possibly just a dodgy textbook question?

Reply 8

The-Spartan

6

Original post by nwmyname

How can I finish this question to find R?

Your working is correct up to the point you solve for

R

.

\dfrac{1}{666.7}=\dfrac{1000+R}{1000R} \iff 1000R=666.7(1000+R)

1000R=666700+666.7R

333.3R=666700

R=\dfrac{666700}{333.3} = 2000.3\Omega

.

As needed.

Reply 9

nwmyname

OP

13

Original post by The-Spartan

Your working is correct up to the point you solve for

R

.

\dfrac{1}{666.7}=\dfrac{1000+R}{1000R} \iff 1000R=666.7(1000+R)

1000R=666700+666.7R

333.3R=666700

R=\dfrac{666700}{333.3} = 2000.3\Omega

.

As needed.

Thank the lord!

Reply 10

uberteknik

21

Original post by nwmyname

Please check the solutions as this is not stated.

Original post by JN17

It looks like you get the correct answer with the assumption that the current is the same, even though it isn't, possibly just a dodgy textbook question?

Original post by The-Spartan

Your working is correct up to the point you solve for

R

.

\dfrac{1}{666.7}=\dfrac{1000+R}{1000R} \iff 1000R=666.7(1000+R)

1000R=666700+666.7R

333.3R=666700

R=\dfrac{666700}{333.3} = 2000.3\Omega

.

As needed.

Original post by nwmyname

Thank the lord!

The Lord has nothing to do with it.

The method Spartan used is correct.

However, the textbook answer to part c) is completely in error.

Moreover, you cannot use the 3mA current calculated in part b), because adding the voltmeter to the circuit will increase the actual current:

You can check the textbook is incorrect by substituting the (textbook) calculated resistance values back into the potential divider circuit which produces a voltage of 2.4V across the voltmeter parallel combination (2K || 1K = 666.7 ohms) and the upper series resistor of 1K.

V_{voltmeter} = V_{supply}(\frac{R_{parallel}}{R_{series} + R_{parallel}})

V_{voltmeter} = 6(\frac{666.7}{1000 + 666.7}) = 2.4V

Ergo this is incorrect and NOT the 2V required by the question.

The error is, as I stated originally, that the text book has used the 3mA unloaded circuit current (i.e. without the voltmeter loading the circuit) when it should have used 4mA:

V_{series} = V_{supply} - V_{parallel} = 6 - 2 = 4V

The total current is then given by

I_{supply} = \frac{V_{series}}{R_{series}} = \frac{4}{1000} = 4 \rm x 10^{-3} A

Correct solution:

R_{parallel} = \frac{V_{parallel}}{I_{supply}}

R_{parallel} = \frac{2}{4 \rm x 10^{-3}} = 500 \Omega

Following through on Spartans working:

\frac{1}{500}=\frac{1000+R}{1000R} \iff 1000R=500(1000+R)

1000R=500000+500R

500R= 500000

R=\frac{500000}{5000} = 1000\Omega

.

Check:

Once again, check this correct working using the potential divider calculation:

R_{parallel} = \rm 1K || 1K = 500\Omega

V_{voltmeter} = V_{supply}(\frac{R_{parallel}}{R_{series} + R_{parallel}})

V_{voltmeter} = 6(\frac{500}{1000 + 500}) = 2V

i.e. the correct p.d. developed across the lower resistor as stated in the question.

(edited 8 years ago)

Reply 11

The-Spartan

6

Original post by uberteknik

The method Spartan used is correct.

Sorry, this is my fault i should have read the question. I only saw the mathematical expressions and jumped in.
After reading, this is infact correct. There is another component and as such there is a different current.
This is the correct response to the question, note @nwmyname i only solved the expression you gave for

R

+1 for correct solution :biggrin:

Physics Electricity Potential Dividers & Resistance [PLEASE HELP]

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you