Voltmeter in series? - Page 1
Forum section: electronics / beginners

author: ace
author_rank: Newbie
timestamp: November 18, 2021, 02:09:15 am
content: Hi, in class we got a question as shown in the picture below, with a circuit where there's a 6V cell in series with 2 6ohm resistors and a voltmeter. and the ammeter is in parallel with one of the 6ohm resistors (the problem says "connected incorrectly by a student"). it said to assume an ideal ammeter, so i think should be measuring 0amps given that there's no resistance. when figuring out the voltage reading of the voltmeter, i figured it would be 0 volts because I believe the 2 6ohm resistors in series can be combined into one 12ohm resistor. Because the two resistors were in series i figured $$R_{total} = 12\Omega$$, and from that, $$I_{total} = 6V / 12\Omega = 0.5A$$, from that, i calculated the voltage drop across the resistor to be $$0.5A * 12\Omega = 6V$$, based off this I would think that the the voltmeter should read 0V, but the answer is listed as 6V. Where did I go wrong?

author: Capernicus
author_rank: Frequent Contributor
timestamp: November 18, 2021, 02:15:57 am
content: it would be 12v on the voltametre, adding a megohm of resistance to the circuit, and the ammetre would read 0 amps, regardless of the voltametre splitting the circuit or not.

author: Tom45
author_rank: Frequent Contributor
timestamp: November 18, 2021, 02:18:45 am
content: An ideal ammeter is 0 ohms. So what is the equivalent resistance of the 6 ohm resistor with an ammeter in parallel with it?An ideal volt meter is an open circuit. For real voltmeters the resistance will be very much greater than any resistors shown in your circuit. So, essentially an open circuit.How much current will flow through each of the resistors given the short circuit of the ammeter and the open circuit of the volt meter?

author: IanB
author_rank: Super Contributor
timestamp: November 18, 2021, 02:19:19 am
content: Quote from: Capernicus on November 18, 2021, 02:15:57 amThe battery supplies 6 V, so 12 V on the voltmeter seems unlikely?

author: IanB
author_rank: Super Contributor
timestamp: November 18, 2021, 02:23:09 am
content: In addition to what Tom45 said, the voltage drop across a resistor is given by \$\Delta V = IR\$Therefore, what is the difference in voltage between each end of a resistor if no current is flowing through it?

author: ace
author_rank: Newbie
timestamp: November 18, 2021, 02:50:17 am
content: Quote from: IanB on November 18, 2021, 02:23:09 amWould it be 0V? $$\Delta V = 0 * 6\Omega = 0V$$

author: IanB
author_rank: Super Contributor
timestamp: November 18, 2021, 02:51:24 am
content: Yes, it would.

author: ace
author_rank: Newbie
timestamp: November 18, 2021, 02:56:57 am
content: Quote from: Tom45 on November 18, 2021, 02:18:45 amWould it be 6 ohms?$${1 \over R_{total} }= {1 \over {6\Omega}} + {1 \over {0\Omega}}$$$$R_{total}= {1 \over {1 \over {6\Omega}} + {1 \over {0\Omega}}} = 6\Omega$$Quote from: Tom45 on November 18, 2021, 02:18:45 amWould it be none because the circuit is open?

author: Cerebus
author_rank: Super Contributor
timestamp: November 18, 2021, 03:03:10 am
content: Quote from: IanB on November 18, 2021, 02:19:19 amNo, no, it's on the Voltametre (which we all know is Alessandro Volta's inside leg measurement). So the probable reading is "Ouch! Where Sir, prey tell, do you think you are going to connect that wire?" (Only in Italian, obviously).

author: IanB
author_rank: Super Contributor
timestamp: November 18, 2021, 03:03:10 am
content: Quote from: ace on November 18, 2021, 02:56:57 amHere's a suggestion for you:$$\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}}=\frac{R_1 R_2}{R_1 + R_2}$$

author: Brumby
author_rank: Unknown
timestamp: November 18, 2021, 03:15:56 am
content: Quote from: ace on November 18, 2021, 02:56:57 amBIG error in evaluating the above!!!What does THIS term resolve to?:$${1 \over {0\Omega}}$$

author: amyk
author_rank: Super Contributor
timestamp: November 18, 2021, 03:16:35 am
content: This is one of those questions where no math should be necessary to answer, only intuition of current and voltage. An ideal voltmeter has an infinite input impedance. From that fact alone, you know no current will flow through the circuit and the voltage shown must be that of the battery, i.e. 6V.(Bonus question for discussion: what if those were 6k resistors? 6M? 6G? 6T?Infinite? What would the voltmeter reading be?)

author: Martian Tech
author_rank: Regular Contributor
timestamp: November 18, 2021, 03:20:02 am
content: Getting back to the original question... You can view it this way:(see attached equivalent circuit. I can't get it to post inline)The resistors in series with the ammeter in parallel with one of them result in some resistance. You don't have to know or care what that is, just that it is small relative to the input impedance of the voltmeter. So you're just measuring the voltage of the battery...

author: Cerebus
author_rank: Super Contributor
timestamp: November 18, 2021, 03:24:15 am
content: Quote from: amyk on November 18, 2021, 03:16:35 amIf the voltmeter input resistance was infinite, and the source resistance infinite, then you'd read infinite voltage from the Johnson noise of the source resistance alone (assuming that measurements are not taken at 0K). However, that voltage would have zero power \$ \frac{V^2}{\infty}\$, and without power to drive the voltmeter, you'd read 0.

author: Brumby
author_rank: Unknown
timestamp: November 18, 2021, 03:25:37 am
content: Quote from: IanB on November 18, 2021, 03:03:10 amThat is a good suggestion.(For those unaware, it is algebraically identical to the original expression)

author: ace
author_rank: Newbie
timestamp: November 18, 2021, 03:45:51 am
content: Quote from: IanB on November 18, 2021, 03:03:10 am$$\frac{R_1 R_2}{R_1 + R_2}= \frac{0\Omega * 6\Omega}{0\Omega + 6\Omega = \frac{0\Omega}{6\Omega} = 0\Omega$$So the total resistance of the the ammeter and 6ohm resistor in parallel is 0 ohms?Quote from: Brumby on November 18, 2021, 03:15:56 am

author: Brumby
author_rank: Unknown
timestamp: November 18, 2021, 04:25:11 am
content: Quote from: ace on November 18, 2021, 03:45:51 amCorrrect.At least when we are talking about textbook exercises - in which one presumesidealcomponents are used.If you want to be exact, things can be a little different when you get into real world - but you MUST understand the ideal component examples in order to get a feel for how things work at a fundamental level. Once you have that under your belt, then allowing for real world situations is reasonably possible.This includes knowing when you can "ignore" real world effects, because they are insignificantfor the case you may be investigating.

author: ace
author_rank: Newbie
timestamp: November 18, 2021, 05:05:14 am
content: So to summarize, in the textbook case there is no drop across the 6ohm resistors because there is no current flowing through the circuit. there is no current flowing through the circuit because the ideal voltmeter has infinite resistance. because there is no voltage drop across the resistors, the voltmeter just reads the 6V of the battery?

author: IanB
author_rank: Super Contributor
timestamp: November 18, 2021, 06:20:28 am
content: Correct.

author: Tom45
author_rank: Frequent Contributor
timestamp: November 18, 2021, 06:25:57 am
content: Quote from: ace on November 18, 2021, 05:05:14 amYes, your answer is correct, and your description of why is also correct.The ammeter across one resistor has no effect on what is happening with the circuit.

author: ledtester
author_rank: Super Contributor
timestamp: November 18, 2021, 09:40:45 am
content: Quote from: IanB on November 18, 2021, 03:03:10 amWhen \$R_1 \ll R_2\$ (\$R_1\$ is much less than \$R_2\$) a useful approximation to remember is that \$ R_1 \parallel R_2 \approx R_1\$ -- i.e. the parallel combination is essentially the smaller of the two resistors.For instance, if \$R_1\$ is one tenth of \$R_2\$, \$ R_1 \parallel R_2 = 0.091 R_2\$ and the approximation gives \$ 0.10 R_2\$.

author: armandine2
author_rank: Frequent Contributor
timestamp: November 18, 2021, 10:41:17 am
content: Exam questions force you to make choices - from OP mine would be1) not an ideal voltmeter -2) resistance of voltmeter, say 10MΩand my calculation using I = V/Rthe ammeter reads I = 6/10 x 106the current in the circuitthe voltmeter reads the battery voltagethus in effect saying neither of the resistors has a measurable affect

author: BeBuLamar
author_rank: Super Contributor
timestamp: November 19, 2021, 12:24:18 am
content: A typical digital voltmeter would read 6V because the 12 Ohms added resistance is nothing compared to the input impedance of the meter of 10 MOhms. The ammeter would read about zero. There is some current goes thru it but very small.

author: Vtile
author_rank: Super Contributor
timestamp: November 19, 2021, 08:24:59 am
content: I think the best answer is symbolical solution answered as function of questioned things, then no one can say that it is de facto incorrect (if no values for Ammeter and Voltmeter is given, especially voltmeter). That is Voltmeter and Ammeter is added to function as resistances and their effect to Voltage and Current is solved out from transfer function. However, this might be too much to ask if necessary mathematical tools are not yet studied in curriculum.If voltmeter is ideal it shows 6 volts (because the internal resistance is infinity and mathematically the effect of all other resistances go to zero) in other hand there is contradiction as the current is zero. Ammeter will have similar contradiction mathematically as the resistance is zero, so basically there is no parallel connection over resistance, but only ammeter in series with one 6 ohms resistance.

author: magic
author_rank: Super Contributor
timestamp: November 19, 2021, 08:34:20 am
content: There is no contradiction.Ideal ammeter is a short circuit.Ideal voltmeter is an open circuit.Modify the schematic like that and you know everything you need. Since there is zero current anywhere, voltage across 6Ω is zero and across the battery there is 6V so that's the voltage the voltmeter sees.

author: Vtile
author_rank: Super Contributor
timestamp: November 19, 2021, 08:43:30 am
content: Quote from: magic on November 19, 2021, 08:34:20 amMy wording is bad, you are correct there is no contradiction in sense of schoolbook idealization.However, there is infinitesimally small error left from other components of the circuit. As of zero and a grain of salt.

author: magic
author_rank: Super Contributor
timestamp: November 19, 2021, 12:52:48 pm
content: Every time you find yourself dealing with a non-negative real number which is less than every positive number, the number you are dealing with is zero exactly, because no other such numbers exist.There is zero current through an ideal voltmeter, by definition.There is zero current through a hypothetical "infinite resistance" too, if by "infinite" you mean "greater than every finite resistance".

author: tooki
author_rank: Super Contributor
timestamp: November 19, 2021, 07:14:55 pm
content: Quote from: ace on November 18, 2021, 02:56:57 amFYI, an easy workaround for div-by-zero problems when calculating resistances is to actually get somewhat closer to real life, wherenothinghas absolutely zero resistance: substitute some minuscule resistance for the zero ohms, like 0.0001 ohms. That allows the math to work, while at the same time being absolutely negligible in the result. In the example above:$$R_{total}= {1 \over {1 \over {6\Omega}} + {1 \over {0.0001\Omega}}} = 0.000099998\Omega$$which you can confidently round down to 0.000 ohms for a school problem.