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Current electricity

  • Thread starter leena19
  • Start date
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1. Homework Statement

You are supplied with a 3V battery and a variable resistor of 30 ohms

http://img12.imageshack.us/img12/8329/currentv.png" [Broken]

a) Complete the above circuit diagram such that it's possible to obtain a PD between 0 and 3 V between 2 terminals.Mark the 2 terminals as A and B(the lines in grey in the figure are the connections I've made)

b) A high resistance voltmeter is now connected between A and B and if its reading is to be 2 V,what should be the resistance to the left and to the right of the setting point of the 30 ohm resistor

c)After adjusting the setting point corresponding to B,a bulb marked 2V,0.6A is connected between A and B,but the bulb won't light up.Explain why

d) It is possible to light the bulb in the normal way by adjusting the resistor.Calculate the approximate resistances to the left and to the right of the resistor after this adjustment is done
2. Homework Equations

[tex]\sum[/tex]E = [tex]\sum[/tex]IR

3. The Attempt at a Solution

b) E = I*30
I = 3/30 = 0.1 A

2 = I*R
2= 0.1R
R= 20 ohms

c)I can't figure out why the bulb doesn't work.When I tried to calculate the resistance of the bulb,I get
R'(of bulb)= 2/0.6 =31/3 ohms which is less than the 20 ohm resistance of the rheostat,so wouldn't most or even all of the current flow through the bulb?If so then why won't it light up?

d) Not sure I can do part d ,without understanding part c),so I hope someone can help.

Thank you.
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 
Last edited by a moderator:

rl.bhat

Homework Helper
4,433
5
c)When you connect the bulb across AB, find the effective resistance across AB?
Then the total Resistance of the circuit and the current in the circuit.
Calculate the potential difference across AB. If it is less than 2 V, the bulb will not light up.
 
186
0
c)When you connect the bulb across AB, find the effective resistance across AB?
Then the total Resistance of the circuit and the current in the circuit.
Calculate the potential difference across AB. If it is less than 2 V, the bulb will not light up.
Thanks for replying.
to find the effective resistance across AB,
1/Rab= 1/20 + 3/10 = 7/20
Rab = 20/7ohms

Rtotal= 20/7 + 10 = 90/7 ohms

cuurent, I, in the circuit
E = IR
3=I*90/7
I=7/30A

the potential difference across AB
Vab=7/30*10/3=7/9V < 2V so the bulb doesn't light up?

Now for part c)
taking the external circuit through the bulb and the potential divider
3 = 0.6(10/3 + R)
R=5-10/3= 5/3 = 1.667ohms from the right?
 

rl.bhat

Homework Helper
4,433
5
Part c is not correct.
If x is the resistance across the bulb, and I is the total current in the circuit, then using branch current formula
0.6 = I*x/(10/3 + x )
I = 0.6*( 10/3 + x )/x..........(1)
When this current flown through the rest of the resistance of the rheostat, potential difference must 3 - 2 = 1 v.
So 1 = I(30 - x) 0r I = 1/ ( 30 - x ).....(2)
Equate eq.(1) and (2) and solve for x.
 
186
0
I'm not familiar with the branch current formula.
Instead,can I use kirchoffs' laws to find x?

3= 2 + I*(30-x) --------(2) and
2= (I-0.6)*x ---------(3)
where I-0.6 = current through x ?

EDIT:
I guess I can't do it this way,but I don't understand why.
Solving for equations (2) and (3),I get x=9.2ohms ?
Solving equations (1) and (2) gives me ,x=0.1975 ohms ?
 
Last edited:

rl.bhat

Homework Helper
4,433
5
I'm not familiar with the branch current formula.
Instead,can I use kirchoffs' laws to find x?

3= 2 + I*(30-x) --------(2) and
2= (I-0.6)*x ---------(3)
where I-0.6 = current through x ?
Yes. You can.
From 2 and 3 find I in terms of x. Equate them and solve for x.
 
186
0
Sorry,I didn't see your reply.
I tried solving it,but I get different answers for both
 

rl.bhat

Homework Helper
4,433
5
Sorry,I didn't see your reply.
I tried solving it,but I get different answers for both
Fro eq.(2) I = 1/( 30 - x )..........(1)
From eq. 3, I = (2 + 0.6x)/x or I = ( 20 + 6x )/10x........(2)
from 1 and 2 we get
1/( 30 - x ) = (20 + 6x )/10x
Now can you solve for x ?
 
186
0
Thanks
After solving
From eq.(2) I = 1/( 30 - x )
From eq. 3, I = (2 + 0.6x)/x or I = ( 20 + 6x )/10x

I get x=28.51 ohms

using branch current formula
0.6 = I*x/(10/3 + x )
I = 0.6*( 10/3 + x )/x..........(1)
I hope by the above equation,you meant
0.6 = I*(10/3+x)/x (which I think is what this site on current divider formula(not sure if this is the same as branch current formula) says
http://www.allaboutcircuits.com/vol_1/chpt_6/3.html" [Broken])

Solving for x using this equation(1) and equation(2) gives x=28.13 ohms and x=0.195ohms,
so would the answer for x be approximately 28 ohms?
 
Last edited by a moderator:

rl.bhat

Homework Helper
4,433
5
When you equate I, you get
1/( 30 - x) = ( 20 + 6x)/10x or
10x = (20+6x)(30-x)
6x^2 - 150x - 600 = 0
x^2 - 25x - 100 = 0
If you solve this quadratic equation, the positive root is x = 28.5 ohm?
How did you ger x = 0.195 ohm?
 
186
0
When you equate I, you get
1/( 30 - x) = ( 20 + 6x)/10x or
10x = (20+6x)(30-x)
6x^2 - 150x - 600 = 0
x^2 - 25x - 100 = 0
If you solve this quadratic equation, the positive root is x = 28.5 ohm?
How did you ger x = 0.195 ohm?
I get x=28.5ohms only when I use those equations,but when I use

0.6 = I*(10/3+x)/x and
I = 1/( 30 - x )
I get 2 answers of x=28.13 ohms and x=0.195 ohms
 

rl.bhat

Homework Helper
4,433
5
I get x=28.5ohms only when I use those equations,but when I use

0.6 = I*(10/3+x)/x and
I = 1/( 30 - x )
I get 2 answers of x=28.13 ohms and x=0.195 ohms
Some thing is wrong. Show the details.
 
186
0
When I use
0.6 = I*[(10/3)+x]/x
I=0.6x/[10/3)+x]
I=18x/100+30x--------------(1)

I=1/(30-x) ----------(2)

(1)=(2)
18x/100+30x = 1/(30-x)
100+30x = 540x-18x2
18x2-540x+30x+100 = 0
18x2-510x+100=0

x=-b[tex]\sqrt[^{+}_{-}]{b^{2}-4ac}/2a[/tex]
x= -(-510)[tex]\sqrt[^{+}_{-}]{(-510^{2})-4*18*100}/2*18[/tex]
x=510)[tex]\sqrt[^{+}_{-}]{260100-7200}/36[/tex]

x=(510+502.89)/36 = 28.13 ohm ?
x=(510-502.89)/36 = 0.1975 ohm ?
 

rl.bhat

Homework Helper
4,433
5
In the formula In = I(total)R(total)/Rn
R(total) is not the sum of the resistances, but the parallel equivalent resistance.
In our problem R(total) = x*(10/3)/( 10/3 + x ).
 
Last edited:
186
0
In the formula In = I(total)R(total)/Rn
R(total) is not the sum of the resistances, but the parallel equivalent resistance.
OK.Then wouldn't R(total) or the equivalent parallel resistance be
1/R(total) = 1/x+3/10
=10+3x/10x

therefore R(total)= 10x/(10+3x) ?

I don't know how to get the below equation :(
In our problem R(total) = x*(10/3 + x )/( 10/3 + 2x ).
 

rl.bhat

Homework Helper
4,433
5
OK.Then wouldn't R(total) or the equivalent parallel resistance be
1/R(total) = 1/x+3/10
=10+3x/10x

therefore R(total)= 10x/(10+3x) ?

I don't know how to get the below equation :(
Yes. You are write.
1/R(total) = 1/x + 1/(10/3)
 
Last edited:
186
0
Then x would be appproximately 29 ohms.

Thanx so much for the help,rl.bhat!
 

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