Potential difference across a resistor Why is potential difference across two ends of a resistor equal to the terminal voltage of the battery?
 A: Electric force is a conservative force (when there is no changing magnetic field). So, the net work done on a charge by electric forces when it returns to the same point after traversing a loop is zero.
Suppose, a charge $q$ falls from the positive terminal of the battery to the negative terminal of the battery, through the battery.
The work done by the electric forces on the charge will be, charge times the negative of potential difference = $-q \cdot [0-(V)] =qV$
Now, suppose, the charge goes to the positive terminal, through the resistor. So, its potential will increase by $=V_{across R})$. So, the work done on the charge will be  $-q[V_{across R}-0]$ = $-qV_{across R}$.
Thus, the net work done on the charge is $qV + [-qV_{across R}]$ $=0$.
From this, it follows that $V_{across R} = V$
From Maxwell's equations, the circulation of the electric field in any closed loop is zero, when there is no changing magnetic field, as in this case. This statement is equivalent to "the electric force is conservative". 

Thus, the line integral of electric field across the resistor from the plus end to minus end (which is the potential difference across two ends of the resistor $=V_{across R})$ plus the line integral of electric field across the terminal of the battery from negative to positive terminal (which is $-V$) must be zero.
So, $V_{across R} + (-V) = 0$
or, $V_{across R} = V$
So, the potential difference across the resistor has to be equal to terminal voltage.
In short, since electric force is conservative, they must be equal.
