RESISTANCE The figure above shows the current-versus-potential-difference graph

Question

RESISTANCE

The figure above shows the current-versus-potential-difference graph for a resistor.

1. What is the resistance of this resistor?

2. Suppose the length of the resistor is doubled while keeping its cross section the same. (This requires doubling the amount of material the resistor is made of.)Drawing on the figure, add to it the current-versus-potential-difference graph for the longer resistor.

You are given a 60 W lightbulb with 240 ? resistance. You calculate the length of the filament to be 60 cm:

1. If the potential difference across the filament is 120 V, what is the strength of the electric field inside the filament?

2. Suppose the length of the bulb?s filament were doubled without changing its diameter or the potential difference across it. What would the electric field strength be in this case?

3. Remembering that the current in the filament is proportional to the electric field, what is the current in the filament following the doubling of its length?

4. What is the resistance of the filament following the doubling of its length?

RESISTORS IN CIRCUITS

1. Three resistors with resistances of 5.0 ?, 9.0 ?, and 12.0 ?, respectively, are connected in a series circuit with a 6.0 V battery. Find the potential difference across the 5.0 ? resistor.

2. Following the steps and answering the questions below, find the equivalent resistance of the combination of resistors shown in the figure:

We will break this complex circuit into simpler combinations of resistors in series or parallel.

a. Identify the two resistors that are in a simple series combination. List them:

i. Lowest: _________

ii. Highest: _________

b. What is the equivalent resistance of the two resistances you listed in part a.?

c. Identify the two resistors that are in a simple parallel combination. List them:

i. Lowest: _________

ii. Highest: _________

d. What is the equivalent resistance of the two resistances you listed in part c.?

We can now replace the two series and two parallel resistances with their equivalent resistances and draw a simpler circuit, as shown above.

e. What is the equivalent resistance of the 8.00 ?? and Rs1 resistors?

We can reduce the diagram even further, as shown below.

f. What is the equivalent resistance of the 1.50 ?? and Rp1 resistors?

Finally, we have reduced the circuit to just two resistors.

g. What is the equivalent resistance of the two resistors, Rp2 and Rs2, above?

Explanation / Answer

(1)
(i)From the given figure we come to know that it is forming straight line and the equation of straight line is Y=mX+ C
and here the value of C is zero.
Therefore Equation becomes Y=mX which will be represent by V= IR in the terms of volatge , current and resistance.
Now to find the value of R . from graph we observe that at V = 5, I is 2.5
Therefore putting the value of V and I in the equation it will give up the value of R = 2 ohm.
(ii) If we double the length, keeping the area constant then the resistance will get doubled by the relation
   R = p * L/A   where p is resistivity which is constant
Now we will find the values for plotting the graph ,
by the relation V= IR and we know that the value of R is 4 ohm which will remain constant
therefore we will vary the value of I to get the different value for V
at I =1 V = 4 , I = 2 V = 8 , I=3 V= 8
now we will join these points after marking on graph which will give us a straight line of slope 1/4 .

(2) (i)NOW , for the second question we have to calcualte the electric field
where dV is potential and dX is length.
We know both the value of potential and length , hence on putting the value will give us the value of elctric filed.
(ii) On doubling the length , the value of resistance will get double. Therefore if we keep the current constant then the value of potential get double which will double the electric filed.
However if we keep the potential constant by decreasing the current to its half then the electric field remain unaffected.
(iii) Same explanation as given above.
(iv) resistance will get double on doubling the length.
Resistors
(i) Since all the resistance are in series therefore the net resistance would be = 5+9+12 = 26
       V= IR , therefore I = 0.23 A , the potential drop across 5 ohm will be = IR = 0.23*5 = 1.15 V
(2) (i) the resistors in series are 3.18 and 3.5   
     (ii) Net Resistance will be = 3.18+3.5 = 6.68
    (iii) The resistances in parallel are 0.75 and 10.2
    (iv) Net equivalent resistances in parallel are Rp1 = (0.75*10.2)/(0.75+10.2) = 0.6986
    (e) Net resistance of 8 and Rs1 similarly as done above 3.64
    (f) The resistnace 1.5 and Rp1 are in series therefore = 0.6986+1.5 = 2.1986
    (g) The Rp2 and Rs2 are in parallel therefore = 1.37 micro ohm

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