$V = IR$ — read aloud: V equals I R — is the chapter's centre of gravity: pressure equals flow times narrowness. Solve it for $V$, for $I$, or for $R$, and one relation answers three exam questions. From here on, every question can ask for any of the three, so the ladder is: pick the formula, rearrange for what you want, put the numbers in their sockets, then compute.

The free check still runs: volts over ohms cancels to amperes, so $I = \dfrac{V}{R}$ survives — and units guide, they don't confess: wrong units always convict a rearrangement; right units never acquit one.

One more thing this chapter loves: the small print. Currents arrive in milliamps and resistances in kilohms all the time. Convert BEFORE the sockets — 20 mA is 0.020 A, and a factor of 1000 lost here comes back as a wrong answer three lines later.