Dont forget the meter loads the circuit when mAking the measurement.

 

Hi Shteii01,

Plug your values back in to the original circuit and you will be able to check if they are correct.

 

Problem 1.

I came up with:
Ra=2558.7 Ohm
Rb=2450.5 Ohm

My answers are very different from yours.

 

My answers are very different from yours.

Yes, looks like mine are wrong. They work for 100 volt, but don't work for 80 volt.

 

They work for 100 volt, but don't work for 80 volt.

The result works out slightly less than 100V.

 

Hint: Ra and Rb happen to be round numbers.

 

Hint: Ra and Rb happen to be round numbers.

Indeed they do.

I found Thevenin equivalents useful in obtaining the solution.

It would be interesting to see how others have correctly solved this. It's not a trivial exercise.

 

Only series- parallel / ohms law used. We dont have Thevenin Theorem yet.

How did you get the value of Ra and Rb..

I used the Voltage Division Theorem is it correct?
in Fig. 1 I use

Vout = Vin ( RL / RL + Ra )

100 V = 220 V ( 10k / 10k + Ra )

Ra = 12k ohms

for Fig. 2

Vout = Vin ( Rb / Rb + 12k )

Vout = 6.857k

is this is correct my answers? THANK YOU SO MUCH SIR/ Engr.

 

Only series- parallel / ohms law used. We dont have Thevenin Theorem yet.

How did you get the value of Ra and Rb..

I used the Voltage Division Theorem is it correct?
in Fig. 1 I use

Vout = Vin ( RL / RL + Ra )

100 V = 220 V ( 10k / 10k + Ra )

Ra = 12k ohms

for Fig. 2

Vout = Vin ( Rb / Rb + 12k )

Vout = 6.857k

is this is correct my answers? THANK YOU SO MUCH SIR/ Engr.

Hi John,

It's encouraging that you have attempted to put down some concrete ideas.

Take your equation for the case of the meter connected across Ra:-

Vout = Vin ( RL / RL + Ra )

Firstly, you need to adopt some good habits concerning writing mathematical expressions ...

I presume you really meant to write:-

Vout = Vin ( RL /( RL + Ra) )

This would still be incorrect. Certainly there is a voltage divider to analyze. The voltage divider is formed as a series resistor combination across the 220V supply. In this configuration one equivalent series resistor is the parallel combination of Ra with the meter 10kΩ loading resistance [Rmeter]. For the moment call this parallel combination Rp_a. The other series element is resistor Rb.

Rp_a would be given by

\(\text{R_{p\_a}=\frac{R_a x R_{meter}}{R_a+R_{meter}}}\)

\(\text{R_{meter}=10k {ohms}}\)

So for the 100V output case, the effective voltmeter reading across the parallel combination Rp_a would be

\(\text{V_{meter}=220\({\frac{R_{p\_a}}{R_{p\_a}+Rb}}\)={100} \ Volts}\)

Can you form a similar equation for the 80V meter reading case with a different equivalent parallel resistance Rp_b? Rp_b would be the parallel combination of Rb with the meter 10kΩ loading resistance [Rmeter]. In this second case, the alternative voltage divider is the series combination of Ra and Rp_b across 220V DC, with the indicated meter voltage being that value formed across the parallel combination Rp_b. You would also need to write the equation for Rp_b.

This is probably enough information for the moment. Give it some thought.

As a further comment I suspect you may struggle to solve this problem without some detailed guidance. Even then you may not readily come to grips with the various principles involved in the overall solution. If you are only just starting out on this type of circuit analysis I think perhaps your professor may be expecting too much of you & your fellow students. How are your fellow students coping with this problem?

 

I found Thevenin equivalents useful in obtaining the solution.

It would be interesting to see how others have correctly solved this. It's not a trivial exercise.

You need to show me how you use Thevenin, because I don't see how it can be use.
I use nodal type analysis and write two simultaneous equation based on KCL.
The first equation look like this
(220V - 120V)/Ra + (220V - 120V)/10k = 120V/Rb
And the second one looks almost the same
(220V - 80V)/Ra = xx + xx
And I also get round nice numbers for Ra = 5K and Rb = xx.

 

Hi Jony,
Thevenin only helps insofar as a short cut to finding the ratio of Ra to Rb. Otherwise not much else. From the point of view of the voltmeter the Thevenin resistance is the same for both cases - Ra||Rb. One still needs to do the required algebraic manipulation - albeit with the advantage of knowing the ratio of Ra to Rb. Hence the problem effectively reduces to one equation with one unknown.

As I indicated previously this isn't a trivial problem.

 

1.Two resistors A and B are connected in series across 220V DC source. When a voltmeter with a internal resistance of 10kΩ is connected across resistor A, the instrument reads 100V and when connected across resistor B, it reads 80V. Find Resistor A and B.

I found Thevenin equivalents useful in obtaining the solution.

It would be interesting to see how others have correctly solved this. It's not a trivial exercise.

I think it's safe to go ahead and solve it.

As usual, there are a number of ways to go about it. Intentionally avoiding a "standard" technique, here is what comes to me off the top of my head.

We know that in the first measurement we have a reading of 100V and so we have a current through the meter of 100V/10kΩ or 10mA. The current resistor A is 100V/Ra while the current in resistor B is 120V/Rb. By KCL, we have

120V/Rb = 100V/Ra + 10mA

We know that in the second measurement we have a reading of 80V and so we have a current through the meter of 80V/10kΩ or 8mA. The current resistor A is 140V/Ra while the current in resistor B is 80V/Rb. By KCL, we have

140V/Ra = 80V/Rb + 8mA

Rearranging these equations we have

140V/Ra - 80V/Rb = 8mA
-100V/Ra + 120V/Rb = 10mA

Multiplying the top equation by 3 and the bottom equation by 2 and then adding the two equations together we get

(3*140V - 2*100V)/Ra = 24mA + 20mA

Ra = 220V/44mA = 5kΩ

 

Hmmm... After reviewing the last part of the thread again, perhaps I jumped the gun too soon. Too late now, since my solution is part of the e-mail notices that went out. Apologies if so.

 

Your solution is almost correct except for a little k.
Nice work anyhow. I did it the long way.

 

Your solution is almost correct except for a little k.
Nice work anyhow. I did it the long way.

Thanks.

When I came up with the first equation, my first reaction was to multiply everything in order to get the resistances upstairs, but that would have left me with an Ra*Rb term. My immediate reaction was, "What the heck is up with getting cross terms? This is a linear system!" Of course, it is not necessarily linear in terms of the resistances, just the voltages and currents. Even so, that got me to thinking about it just enough to realize that I had a linear equation in (1/Ra) and (1/Rb), or the conductances. After that, it was a cake walk.

 

thanks i gotta study how did you solve that. then i will work with my new solution using the standard formula. Thank you so much sir/Engr.

Too pro!

 

I think it's safe to go ahead and solve it.

As usual, there are a number of ways to go about it. Intentionally avoiding a "standard" technique, here is what comes to me off the top of my head.

We know that in the first measurement we have a reading of 100V and so we have a current through the meter of 100V/10kΩ or 10mA. The current resistor A is 100V/Ra while the current in resistor B is 120V/Rb. By KCL, we have

120V/Rb = 100V/Ra + 10mA

We know that in the second measurement we have a reading of 80V and so we have a current through the meter of 80V/10kΩ or 8mA. The current resistor A is 140V/Ra while the current in resistor B is 80V/Rb. By KCL, we have

140V/Ra = 80V/Rb + 8mA

Rearranging these equations we have

140V/Ra - 80V/Rb = 8mA
-100V/Ra + 120V/Rb = 10mA

Multiplying the top equation by 3 and the bottom equation by 2 and then adding the two equations together we get

(3*140V - 2*100V)/Ra = 24mA + 20mA

Ra = 220V/44mA = 5kΩ

where the -80/Rb + 120/Rb>

 

Do you know how to solve simultaneous equations?

 

where the -80/Rb + 120/Rb>

Multiply top equation by 3, that gives you -240V/Rb in that equation.

Multiply the bottom equation by 2, that gives you +240V/Rb in that equation.

Add them up and they go away.