# Homework Help: Current electricity

1. Jul 22, 2009

### leena19

1. The problem statement, all variables and given/known data

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. Relevant equations

$$\sum$$E = $$\sum$$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. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 4, 2017
2. Jul 22, 2009

### rl.bhat

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.

3. Jul 25, 2009

### leena19

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?

4. Jul 25, 2009

### rl.bhat

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.

5. Jul 26, 2009

### leena19

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: Jul 26, 2009
6. Jul 26, 2009

### rl.bhat

Yes. You can.
From 2 and 3 find I in terms of x. Equate them and solve for x.

7. Jul 26, 2009

### leena19

I tried solving it,but I get different answers for both

8. Jul 26, 2009

### rl.bhat

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 ?

9. Jul 26, 2009

### leena19

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

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

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: May 4, 2017
10. Jul 26, 2009

### rl.bhat

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?

11. Jul 26, 2009

### leena19

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

12. Jul 26, 2009

### rl.bhat

Some thing is wrong. Show the details.

13. Jul 26, 2009

### leena19

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

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

14. Jul 26, 2009

### rl.bhat

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: Jul 27, 2009
15. Jul 26, 2009

### leena19

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 :(

16. Jul 27, 2009

### rl.bhat

Yes. You are write.
1/R(total) = 1/x + 1/(10/3)

Last edited: Jul 27, 2009
17. Jul 27, 2009

### leena19

Then x would be appproximately 29 ohms.

Thanx so much for the help,rl.bhat!