Current Electricity Homework: 3V Battery & 30 Ohm Resistor

In summary, the conversation discusses a circuit with a 3V battery and a variable resistor of 30 ohms. The circuit is completed by adding a high resistance voltmeter and a bulb. In order to obtain a PD between 0 and 3 V, the circuit is adjusted by calculating the effective resistance across AB. The bulb does not light up due to the potential difference across AB being less than 2V. Using Kirchoff's laws, the approximate resistances to the left and right of the resistor can be calculated.
  • #1
leena19
186
0

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

Homework Equations



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

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.
 
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  • #2
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
rl.bhat said:
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?
 
  • #4
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
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 ?
 
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  • #6
leena19 said:
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.
 
  • #7
Sorry,I didn't see your reply.
I tried solving it,but I get different answers for both
 
  • #8
leena19 said:
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 ?
 
  • #9
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?
 
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  • #10
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
rl.bhat said:
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
 
  • #12
leena19 said:
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.
 
  • #13
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 ?
 
  • #14
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 ).
 
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  • #15
rl.bhat said:
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 :(
rl.bhat said:
In our problem R(total) = x*(10/3 + x )/( 10/3 + 2x ).
 
  • #16
leena19 said:
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)
 
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  • #17
Then x would be appproximately 29 ohms.

Thanx so much for the help,rl.bhat!
 

1. What is the purpose of using a 3V battery and a 30 Ohm resistor in this homework?

The purpose of using a 3V battery and a 30 Ohm resistor is to demonstrate the application of Ohm's Law, which states that the current through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them. By using a specific voltage and resistance, we can calculate the current flowing through the circuit and understand the relationship between these three variables.

2. How do I calculate the current through the circuit using the given values?

To calculate the current through the circuit, simply use Ohm's Law formula: I = V/R. In this case, the voltage (V) is 3V and the resistance (R) is 30 Ohms. Therefore, the current (I) flowing through the circuit is 0.1A or 100mA.

3. Can I use different values for the battery and resistor and still get the same result?

No, using different values for the battery and resistor will result in a different current flowing through the circuit. The current is directly proportional to the voltage and inversely proportional to the resistance, so changing either of these values will affect the overall current in the circuit.

4. How does the current change if I add another 30 Ohm resistor in series with the existing resistor?

If you add another 30 Ohm resistor in series with the existing resistor, the total resistance in the circuit will be 60 Ohms. This will cause the current to decrease, as the resistance has doubled while the voltage remains the same. The new current can be calculated using Ohm's Law: I = V/R = 3V/60 Ohms = 0.05A or 50mA.

5. What happens to the current if I increase the voltage of the battery to 6V?

If you increase the voltage of the battery to 6V, the current will also increase. This is because the voltage and current have a direct relationship, meaning that as one increases, the other also increases. The new current can be calculated using Ohm's Law: I = V/R = 6V/30 Ohms = 0.2A or 200mA.

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