Electric Circuits

[Zingerburger7]

In the circuit attached, calculate the battery voltage.

 

I know that V=IR, but I need help in figuring out how to use the 4 A. For the total voltage, its the sum of all of the resistance in the circuit multiply the current, but I don't know how to find out the current.

Thanks.

[ewmon]

Try applying Kirchhoff's laws.

[TonyMcC]

Homework? If it is then show us what you have tried. Actually I think Ohm's Law and "Resistors in series and parallel" is all you need.

Edited July 8, 2011 by TonyMcC

[Zingerburger7]

um I've tried to find the total resistance, which I think you get by adding the series (2 and 4) and the parallels (8, 16 and 16). Then multiply by the current which is 4, but the answer I got was wrong. Maybe I need to do something to the 4 A? Or did I add my resistance up wrong (I got 10 Ohms). It's not a hard question (the textbook says so) but I just can't get my head around it.

Thanks again.

[ewmon]

I actually haven't worked on such problems in a while, but the rightmost leg contains a current source (4A) and a resistance (16Ω). Next step (I think): Use a simple formula to simplify the circuit.

[DrRocket]

I actually haven't worked on such problems in a while, but the rightmost leg contains a current source (4A) and a resistance (16Ω). Next step (I think): Use a simple formula to simplify the circuit.

 

The circuit is linear. Write and solve equations for loop or nodal currents (not a mixture of both) using Kirchoff's laws. There will be several simultaneos linear equations to be solved.

 

If this is for a physics class, use nodal currents since the loop current method is not usually taught in physics courses.

 

But in any case, more information is needed. Both the voltage source and the battery voltage could be anything as the circuit is given.

Edited July 9, 2011 by DrRocket

[TonyMcC]

Note the 8, 16 and 16 ohm resistors are in parallel and so have the same voltage across them (V).

 

This voltage can be calculated from knowing 4A flows through one of the 16 ohm resistors and is 64V (V=IR).

 

The 8,16 and 16 ohm network has an effective resistance of 4 ohms (1/Rt= 1/R1 +1/R2 + 1/R3).

 

An equivalent circuit can therefore be considered consisting of 2, 4 and 4 ohms in series with 64V across one of the 4 ohm resistors.

 

Since all resistors in a series circuit carry the same current, the voltage across them is proportional to resistance.

 

Therefore total voltage around the circuit is 32+64+64= 160V. (IMO).

 

The battery voltage is therefore 160V

 

um I've tried to find the total resistance, which I think you get by adding the series (2 and 4) and the parallels (8, 16 and 16). Then multiply by the current which is 4, but the answer I got was wrong. Maybe I need to do something to the 4 A? Or did I add my resistance up wrong (I got 10 Ohms). It's not a hard question (the textbook says so) but I just can't get my head around it.

Thanks again.

 

You are correct in thinking the current supplied by the battery is not 4A.

The battery supplies 4A through both both 16 ohm resistors and 8A through the 8 ohm resistor. Since these resistors are in parallel the battery supplies 16A. If you use 16A I believe you will get the correct answer.

Edited July 9, 2011 by TonyMcC

[J.C.MacSwell]

Note the 8, 16 and 16 ohm resistors are in parallel and so have the same voltage across them (V).

 

This voltage can be calculated from knowing 4A flows through one of the 16 ohm resistors and is 64V (V=IR).

 

The 8,16 and 16 ohm network has an effective resistance of 4 ohms (1/Rt= 1/R1 +1/R2 + 1/R3).

 

An equivalent circuit can therefore be considered consisting of 2, 4 and 4 ohms in series with 64V across one of the 4 ohm resistors.

 

Since all resistors in a series circuit carry the same current, the voltage across them is proportional to resistance.

 

Therefore total voltage around the circuit is 32+64+64= 160V. (IMO).

 

The battery voltage is therefore 160V

 

 

 

You are correct in thinking the current supplied by the battery is not 4A.

The battery supplies 4A through both both 16 ohm resistors and 8A through the 8 ohm resistor. Since these resistors are in parallel the battery supplies 16A. If you use 16A I believe you will get the correct answer.

 

You do know that but the 4A is a current "source", not a meter reading. Correct? So it is a variable voltage source as well. (it will give a potential required to make it 4A)

Edited July 9, 2011 by J.C.MacSwell

[John Cuthber]

I don't think it's a source, I think it's a meter.

It has the same format as the other meter (the voltmeter)

and also, if it's a source the problem is not soluble because you don't know the battery voltage.

[J.C.MacSwell]

I don't think it's a source, I think it's a meter.

It has the same format as the other meter (the voltmeter)

and also, if it's a source the problem is not soluble because you don't know the battery voltage.

 

OK, I assumed that was a voltage source also. So what do you make of the circuit connecting the top and bottom at the voltmeter? Is that actually part of the circuit? Or is it just the connections for the voltmeter?

 

If it is part of the circuit with no resistance then what drives the 4A?

[Hal.]

In the circuit attached, calculate the battery voltage.

 

I know that V=IR, but I need help in figuring out how to use the 4 A. For the total voltage, its the sum of all of the resistance in the circuit multiply the current, but I don't know how to find out the current.

Thanks.

 

 

 

4 amps flow through the 16 ohm resistance that is on the far right . This is in parallel with another 16 ohm resistance . Both paths provide equal opposition to electric current , so , the current through each of them is 4 amps .

 

Just to the left of the ammeter there is a junction . The current entering this junction is equal to the current leaving this junction . The current leaving is 8 amps , so 8 amps has entered .

 

Directly below this junction there is a point named z which is above another junction . Any current entering into the top junction is equal to any current leaving the bottom junction .

 

So , where there is a voltmeter shown there is a current of 8 amps flowing past one of it's terminals and 8 amps flowing past the other of it's terminals .

 

The two 16 ohm resistances are the equal of one 8 ohm resistance . This equal resistance is in parallel with the 8 ohm resistance that is in the loop on the far left . They provide equal opposition to the flow of current .

 

So , there is 8 amps flowing in one of the directions away from the junction that is to the right of the 2 ohm resistance and there is 8 amps flowing from the other option .

 

16 amps entered the junction and 16 amps joins back together again at the junction below the point y .

 

I think the currents are known in each part of the loop on the far left , now . There is 16 amps through the 2 ohm resistance , 8 amps through the 8 ohm resistance and 16 amps through the 4 ohm resistance .

 

The volt drop across a resistance is equal to the multiplication of the current and the resistance . Across the 2 ohm resistance this is 32 volts , across the 8 ohm resistance this is 64 volts and across the 4 ohm resistance this is 64 volts .

 

The total volt drop in this loop is 160 volts . The sum of the volt drops is equal to the sum of the volt rises . The power source is therefore 160 volts .

 

What I have done in going from one part to the next using mental arithmetic is easy in this case and this approach can be extremely difficult to keep track of in more complicated circuits .

 

Hal_2011

[TonyMcC]

Firstly the circuit symbol for a voltmeter is a circle with a V inside and for an ammeter is a circle with an A inside so the circuit contains a couple of meters.

 

Secondly there is nothing wrong with Hal's answer.

 

These problems are a bit like the Windows operating system in that there is usually more than one way of getting what you want.

 

I feel my approach (#7) is, if anything, the more simple approach and pleased that both Hal and I agree on the answer.

 

In answer to a point asked earlier, in circuit diagrams the lines in a circuit connecting components are assumed to have no resistance.

[Hal.]

I think it is important for people to realise this is an ideal situation . I wouldn't build such a circuit and use it unless someone like TonyMcC gave his approval . We are talking about taking 16 amps from a 160 volt supply , 2560 Watts . That's dangerous .

 

 

[TonyMcC]

OK, I assumed that was a voltage source also. So what do you make of the circuit connecting the top and bottom at the voltmeter? Is that actually part of the circuit? Or is it just the connections for the voltmeter?

 

If it is part of the circuit with no resistance then what drives the 4A?

 

The 160V battery drives all of the current.

Going clockwise from the battery, all the current passes through the 2 ohm resistor. It then splits into three paths as it passes through the 8 ohm and the two 16 ohm resistors. It then recombines to pass through the 4 ohm resistor.

 

I think it is important for people to realise this is an ideal situation . I wouldn't build such a circuit and use it unless someone like TonyMcC gave his approval . We are talking about taking 16 amps from a 160 volt supply , 2560 Watts . That's dangerous .

 

That is why I wondered if it was a homework question! It is the sort of (silly?) theoretical question that a lecturer might put to his students to test their understanding of DC circuit theory. It might not make much practical sense - but I have to plead guilty to doing this myself in the past - lol.

Edited July 9, 2011 by TonyMcC

[ewmon]

I actually haven't worked on such problems in a while, but the rightmost leg contains a current source (4A) ...

Yeah, oops. Thanks! It's an ammeter. I haven't worked on such problems for "quite a while". Now it makes a lot more sense.

[Zingerburger7]

Thanks guys!

[J.C.MacSwell]

Firstly the circuit symbol for a voltmeter is a circle with a V inside and for an ammeter is a circle with an A inside so the circuit contains a couple of meters.

 

Secondly there is nothing wrong with Hal's answer.

 

These problems are a bit like the Windows operating system in that there is usually more than one way of getting what you want.

 

I feel my approach (#7) is, if anything, the more simple approach and pleased that both Hal and I agree on the answer.

 

In answer to a point asked earlier, in circuit diagrams the lines in a circuit connecting components are assumed to have no resistance.

 

I knew the lines had no resistance.

 

The voltmeter is extraneous in the set up. It tells you nothing. I assumed it was a voltage source and the ammeter a current source.

 

What you (and Hal) did looks correct...assuming no resistance at the ammeter and infinite resistance at the voltmeter.

[Hal.]

From the Wiki about a Voltmeter ! source : http://en.wikipedia.org/wiki/Voltmeter

 

A moving coil galvanometer can be used as a voltmeter by inserting a resistor in series with the instrument. It employs a small coil of fine wire suspended in a strong magnetic field. When an electric current is applied, the galvanometer's indicator rotates and compresses a small spring. The angular rotation is proportional to the current through the coil. For use as a voltmeter, a series resistance is added so that the angular rotation becomes proportional to the applied voltage.

 

One of the design objectives of the instrument is to disturb the circuit as little as possible and so the instrument should draw a minimum of current to operate. This is achieved by using a sensitive ammeter or microammeter in series with a high resistance.

 

The sensitivity of such a meter can be expressed as "ohms per volt", the number of ohms resistance in the meter circuit divided by the full scale measured value. For example a meter with a sensitivity of 1000 ohms per volt would draw 1 milliampere at full scale voltage; if the full scale was 200 volts, the resistance at the instrument's terminals would be 200,000 ohms and at full scale the meter would draw 1 milliampere from the circuit under test. For multi-range instruments, the input resistance varies as the instrument is switched to different ranges.

 

Moving-coil instruments with a permanent-magnet field respond only to direct current. Measurement of AC voltage requires a rectifier in the circuit so that the coil deflects in only one direction. Moving-coil instruments are also made with the zero position in the middle of the scale instead of at one end;these are useful if the voltage reverses its polarity.

 

Voltmeters operating on the electrostatic principle use the mutual repulsion between two charged plates to deflect a pointer attached to a spring. Meters of this type draw negligible current but are sensitive to voltages over about 100 volts and work with either alternating or direct current.

 

From the Wiki about an Ammeter ! source : http://en.wikipedia.org/wiki/Ammeter

 

The majority of ammeters are either connected in series with the circuit carrying the current to be measured (for small fractional amperes), or have their shunt resistors connected similarly in series. In either case, the current passes through the meter or (mostly) through its shunt. They must not be connected to a source of voltage; they are designed for minimal burden, which refers to the voltage drop across the ammeter, which is typically a small fraction of a volt. They are almost a short circuit.

 

Ordinary Weston-type meter movements can measure only milliamperes at most, because the springs and practical coils can carry only limited currents. To measure larger currents, a resistor called a shunt is placed in parallel with the meter. The resistances of shunts is in the integer to fractional milliohm range. Nearly all of the current flows through the shunt, and only a small fraction flows through the meter. This allows the meter to measure large currents. Traditionally, the meter used with a shunt has a full-scale deflection (FSD) of 50 mV, so shunts are typically designed to produce a voltage drop of 50 mV when carrying their full rated current.

 

Zero-center ammeters are used for applications requiring current to be measured with both polarities, common in scientific and industrial equipment. Zero-center ammeters are also commonly placed in series with a battery. In this application, the charging of the battery deflects the needle to one side of the scale (commonly, the right side) and the discharging of the battery deflects the needle to the other side. A special type of zero-center ammeter for testing high currents in cars and trucks has a pivoted bar magnet that moves the pointer, and a fixed bar magnet to keep the pointer centered with no current. The magnetic field around the wire carrying current to be measured deflects the moving magnet.

 

Since the ammeter shunt has a very low resistance, mistakenly wiring the ammeter in parallel with a voltage source will cause a short circuit, at best blowing a fuse, possibly damaging the instrument and wiring, and exposing an observer to injury.

 

In AC circuits, a current transformer converts the magnetic field around a conductor into a small AC current, typically either 1 A or 5 A at full rated current, that can be easily read by a meter. In a similar way, accurate AC/DC non-contact ammeters have been constructed using Hall effect magnetic field sensors. A portable hand-held clamp-on ammeter is a common tool for maintenance of industrial and commercial electrical equipment, which is temporarily clipped over a wire to measure current. Some recent types have a parallel pair of magnetically-soft probes that are placed on either side of the conductor.