In one circuit, the terminals are at 98V and 100V. In the second circuit, they are at 2V and 6V. Why does the 2V and 6V produce a stronger electric field?

The potential difference between circuit 1 = 100V - 98V

                                                                      = 2V

The potential difference between circuit 2 = 6V - 2V

                                                                       = 4V

According to Ohm's law,

ΔV = RI

where,

ΔV is the potential difference

R is the resistance

I is the current

We can see that potential difference is directly proportional to the current i.e., greater the potential difference, greater will be the current and thus more will be the electric field.

And as per the potential difference of the two circuits, circuit 2 has greater potential difference than the circuit 1.

So, the circuit 2 will have greater produce stronger electric field.

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.