2ohm and 4ohm resistor are connected in series to a 12V battery what is the current through each resistor and use Ohm's law 2 show that the voltage drops across the individual resistors add up to 12V?

Answer 1

To answer this question, you need not only Ohm's law, but also Kirchoff's current and voltage Laws.

Kirchoff's current law say the current at every point in a series circuit is the same. Lets call that current i.

Ohm's law say voltage is resistance times current. That means the voltage across each resistor R1 and R2 is V1 = R1i and V2 = R2i.

Kirchoff's voltage law says that the signed voltage drops around a series circuit add op to zero. This means that the voltages V1 and V2 must add up to be be equal to 12. This means that R1i + R2i = 12. This means that i = 12/(R1 + R2). But, hey, you say, this is just Ohm's law! This means that the current through the circuit is 2A. It also means that the resistance of two resistors in series is simply the sum of their resistance.
Now that we know the current, we can calculate each resistor's voltage. V1 = R1i = (2)(2) = 4, and V2 = R2i = (4)(2) = 8. Crosscheck 4 + 8 = 12, so the two voltages do add up to twelve volts, as expected.

In summary: The current through both resistors is 2A. The voltage across the 2 ohm resistor is 4V. The voltage across the 4 ohm resistor s 8V.