# Kirchoff Loop and current

The pic of the circuit/problem:

I got the problem right, but i want to make sure my reasoning is correct for the current left of the 2.1 V battery. Its the same current as $I_1$ because the charge carriers do not split in anyway, and the battery only does work on them moving the carriers from a low potential to a high potential. However, if $I= \frac{dQ}{dt}$, wouldn't supplying energy via moving the charges from low potential to high potential cause the charges to move more quickly, i.e. increase the current?

A little fuzzy on why it is the same current, if someone could enlighten me I would very much so appreciate it.

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Same current as what? $I_1$? At any rate, if something increased the current to the left of the battery, those carriers have to come from somewhere: the right of the battery. The currents would change in lock-step. –  garyp Mar 5 at 3:33