Series circuit questions

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zappy

Senior Member
Location
CA.
Question 1. The total Voltage is 75v. The first resistor is 1k the next one is 2k. What is the voltage drop across the 2k resistor? I need to know how you came up with the answer please.

Question 2. The total voltage in the circuit is 10v. there are 3 resistors, one is 50 ohms, second is 10 ohms, third is 40 ohms. what is the power dissipated by the 10 ohm resistor. Please explain how you got the answer.

Question 3. you have two batteries in series, one is 20v. second is 30v. you have three resistors, one is 50 ohms, second is 100 ohms third is 50 ohms. What is the current in the circuit? Please explain how you got the answer.

I keep trying to use ohms law, but only knowing the voltage and ohms in the circuit , I cant figure it out:-?
 

nakulak

Senior Member
did you pay attention in class, or skip class ?

V=IR. in series, resistances add. create yourself an equivalent circuit.
so in example 1, the equivalent ckt is 75 volts accross 3ohms. solve for I (current). I is same though both resistors since they are in series, solve for E for each resistor to get the individual voltage drops.
 

zappy

Senior Member
Location
CA.
did you pay attention in class, or skip class ?

V=IR. in series, resistances add. create yourself an equivalent circuit.
so in example 1, the equivalent ckt is 75 volts accross 3ohms. solve for I (current). I is same though both resistors since they are in series, solve for E for each resistor to get the individual voltage drops.
Question 1. 75v. divided by 3k =.025 amps x 2k = 50v. Ok that's what I was doing wrong, I wasn't adding up all the resistors to get the amps. for the whole circuit. Thank you.
 

zappy

Senior Member
Location
CA.
I generally draw out the ckt. It makes it a lot easier for me to see it - like this.

cf
Thanks for the drawing. The answer for question 1. is 25v. because it was 75v.total and I got 50v. from the equation so it dropped 25v. is this correct?

Question 2. I got 1 watt is that correct?

Question 3. I got 0.25amps is that correct? Thank you.
 

Cold Fusion

Senior Member
Location
way north
quote=zappy;1069721

The answer for question 1. is 25v. because it was 75v.total and I got 50v. from the equation so it dropped 25v. is this correct?
No, it's 50V. When they ask for a "voltage drop" they want the voltage you would measure if you put a voltmeter across the resistor. So I x R is the voltage one would measure across the resistor which is 50V

Question 2. I got 1 watt is that correct?
I got 0.1W. one of us missed a decimal point

Question 3. I got 0.25amps is that correct?
That's what I got. 50volts/200ohms = 0.25A

cf
 

zappy

Senior Member
Location
CA.
Ok on question 1. Why does the 2k have 50v. and the 1k have 25v.? Shouldn't the more resistance lower the voltage? On question 2. 10v. divided by 100 ohms = 0.1amps, then x that by ten gives you 1 watt being dissipated by the 10 ohm resister correct?
 

Cold Fusion

Senior Member
Location
way north
--- On question 2. 10v. divided by 100 ohms = 0.1amps, then x that by ten gives you 1 watt being dissipated by the 10 ohm resister correct?
I x R = E (That's a translation of Mr Ohm's paper :smile:)

(1) P = E x I
(2) P= (I^2) x R
(3) P = (E^2)/R

you can derive (2) and (3) from (1) and substituting in from ohms law

The one you and I used is (2):

P = (.1) x (.1) x 10 = .1W

I don't have a good answer for the first Q. I'll have to thinkabout how to explain that one.

cf
 

Cold Fusion

Senior Member
Location
way north
Ok on question 1. Why does the 2k have 50v. and the 1k have 25v.? ...
For the wizards out there: This is limited to DC and pure resistnace. (anyone else that wants to pitch in here won't hurt my feelings)

Okay, a little circuit theory. I think you are just missing what some of the terms mean.

We got Ohm E = I x R That gives the relation for voltage and current for a resistor. If you know the current through a resistor and the resistance, then you can calculate the voltage across the resistor.

So the current around the loop is: I = (75V)/(3000ohms) = .025A

So, if you put your meter probes across the 1000ohm resistor what is the voltage? E = I x R = .025 x 1000 = 25V

So, if you put your meter probes across the 2000ohm resistor what is the voltage? E = I x R = .025 x 2000 = 50V.

--- Shouldn't the more resistance lower the voltage? ...
I'm not sure what you are asking. More resistance give less voltage drop? Well, no.

Voltage across a resistor = I x R (right?)

In this case, both reisitors have the same current through them (right?)

So with the current the same, the highe the resistor, the higher voltage drop.
.025 x 1000 is less than .025 x 2000

Draw the circuit. Label all the compoents, values, current through them. Calculate the Voltages. See what you get.

cf
 

zappy

Senior Member
Location
CA.
I x R = E (That's a translation of Mr Ohm's paper :smile:)

(1) P = E x I
(2) P= (I^2) x R
(3) P = (E^2)/R

you can derive (2) and (3) from (1) and substituting in from ohms law

The one you and I used is (2):

P = (.1) x (.1) x 10 = .1W

I don't have a good answer for the first Q. I'll have to thinkabout how to explain that one.

cf
Thank you for all your help! So you have to use P = (I^2)X R to get the correct answer? Because when I use P = (E X I), I get 1 watt not .1 watt and when I use P = (E^2)/R I get 1 watt also.
 

Cold Fusion

Senior Member
Location
way north
Thank you for all your help! So you have to use P = (I^2)X R to get the correct answer? Because when I use P = (E X I), I get 1 watt not .1 watt and when I use P = (E^2)/R I get 1 watt also.
Check your math. The voltage across the 10ohm resistor E(r10) = I x R = .1 x 10 - 1V

cf
 

zappy

Senior Member
Location
CA.
Check your math. The voltage across the 10ohm resistor E(r10) = I x R = .1 x 10 - 1V

cf
Huh? .1 x 10 = 1 watt is that what you meant to say? So are you saying the correct answer is 1 watt or 0.1 watt?:-? Your probley sick of me by now:grin:
 

dbuckley

Senior Member
The first example does not need Ohms Law to solve: it is a classical potetnial divider and can be solved with simple proportions. The ratio of the resistances is 2:1 and therefore the ratio of the voltages is 2:1. So 2+1 = 3, and strangely 75V divides nicely by three...

The second example also features a potential divider. You've got three resistors in series that conveniently add up to 100 ohms, and 10V across them. Using the formula wheel you memorised (look for it on Google image search and print one out), P=E^2/R. so P=10^2/100 = 100/100 = 1 watt total dissapation of power. So, again using proportions, every ten ohms dissapates a tenth of a watt.

The last is straight Ohms law: nothing complex about that example.
 

zappy

Senior Member
Location
CA.
The first example does not need Ohms Law to solve: it is a classical potetnial divider and can be solved with simple proportions. The ratio of the resistances is 2:1 and therefore the ratio of the voltages is 2:1. So 2+1 = 3, and strangely 75V divides nicely by three...

The second example also features a potential divider. You've got three resistors in series that conveniently add up to 100 ohms, and 10V across them. Using the formula wheel you memorised (look for it on Google image search and print one out), P=E^2/R. so P=10^2/100 = 100/100 = 1 watt total dissapation of power. So, again using proportions, every ten ohms dissapates a tenth of a watt.

The last is straight Ohms law: nothing complex about that example.
I finally see what I was doing wrong. Thanks for your help too.
 

Cold Fusion

Senior Member
Location
way north
Huh? .1 x 10 = 1 watt is that what you meant to say? ---
No, I said it right. I'll put the units in maybe that will help.
Check your math. The voltage across the 10ohm resistor E(r10) = I x R = .1A x 10ohms = 1V
E(r10) means, "the voltage across the 10ohm resistor"

Repeat after me:
Ohms Law E = I x R

Somebody else's Law about Power
I squared times R is power
E x I is power
E squared divided by R is power

cf
 

USMC1302

Senior Member
Location
NW Indiana
Zappy: If I remember way back, some texts used to give an explanation of voltage drop or potential difference as "electrical pressure". If we think about it in that sense, in a series circuit where the current flow will be constant through each resistive load, it would make sense to think that for a larger resistance, more "pressure" would be required or a larger voltage drop would occur? Sure sounded good in my head.
 
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