電動勢難題高手請進(20分)

I suppose the graph given in your question can be represented by a battery of emf (electromotive force) E, with internal resistance r, connected to an external variable resistance R.

The voltage across the external variable resistance R is U (as given by the y-axis of the graph) when the current through it is I.

Apparently, the graph can be represented by the equation,
U = E - (R+r).I
When R is large as compared to r, the curve is a decreasing straight line.

When R is very very large (i.e. open circuit condition), almost all of the emf of the battery appears across R, and the current is practically zero (I=0 on the graph). Hence, the y-intercept would give the emf of the battery.

When R is adjusted to a zero value (short circuit condition), there is no voltage across it U=0 on the graph). All emf of the battery would then appear across its internal resistance r. Thus the x-intercept gives the current under this condition.

With the above understanding, then...

Option (A): The short circuit current is given by the value at the x-intercept, it is clear that I1 > I2
Option (B): Internal resistance is given by the emf/short circuit current, i.e. the slope of the line. It is thus clear that r2 > r1
Option (C), if the circuit is unchanged, increasing the emf of , say E1, means shifting the line upward. But the slope of which still remain unchanged. Thus r2>r1 still holds
Option (D): since the slope of "line 2" is steeper than that of "line 1", a unit change of current leads to a larger than in U in "line 2".

電動勢E1=E2 , 發生短路時的電流 I 1>I 2
B 內阻r 1>r 2
C 電動勢E1>E2 , 內阻r 1<r 2
D 當兩個電源的工作電流變化量相同時,
電源2的路端電壓變化較電源1的大