Why Does the Internal Resistance Graph Have a Negative Gradient?

Click For Summary
The internal resistance graph exhibits a negative gradient due to the relationship between voltage and current, where internal resistance is factored in. For ohmic resistors, the graph shows a positive gradient, but when considering internal resistance, the equation E = I(R + r) reveals that V = E - Ir, indicating a negative coefficient for current. This means that at V=0, the current is not at its maximum, contradicting some assumptions. The negative gradient of the line is represented by -r, with E as the y-intercept. Understanding this relationship clarifies the behavior of circuits with internal resistance.
Vaseline
Messages
4
Reaction score
0
When taking internal resistance into account, voltage/current graph has negative gradient (i.e. internal resistance).?

My thoughts...

Because for ohmic resistors, graph has a positive gradient. Why does one say that when V=0, Current = max, whilst the other says that when V=0, Current = 0?
 
Physics news on Phys.org
Vaseline said:
When taking internal resistance into account, voltage/current graph has negative gradient (i.e. internal resistance).?

My thoughts...

Because for ohmic resistors, graph has a positive gradient. Why does one say that when V=0, Current = max, whilst the other says that when V=0, Current = 0?

My question was answered in that thread, this is the result I got.

E = I(R+r) where R is the total resistance in the circuit and r is the internal resistance of the cell or battery. From this we can see that,

E = V + Ir

and,

V = E - Ir

When plotting a graph of potential different against current it is clear that the current I has a negative co-efficient: -r. Hence -r is the gradient of the line, and E is the y intercept. y=mx +c (remember?) this is just V=-rI+E.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
12
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
9
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 19 ·
Replies
19
Views
2K
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K