Questions
Question Why is the resistance of the portion of the nichrome wire equal to the resistance of whatever device is connected between the contacts a and b when the identical comparison bulbs are of equal brightness?
Given P_1=P_2 (equal brightness for the comparison bulbs),
then (I_1)(V_1)=(I_2)(V_2) .
For the device I=I(V) or V=V(I), and thus
I_1=I_2 and V_1=V_2 .
By Kirchhoff's voltage law applied to the two branches
V_ba=V_1 + V_device
V_ba=V_2 + V_N .
Thus
V_device = V_N
and thus
R_device =V_device/I_1 = V_N/I_2 = R_N
and thus
R_device=R_N .
That completes the demonstration.
In questions like this, students frequently just write down correct elements of the answer, but fail to make the elements cohere into a logically correct followable demonstration. They often resort to non sequiturs.