So = 4.
WHAT IS THE CURRENT RUNNING THROUGH R1 AND R2?
We know that the total voltage drop is 12 V, and since the voltage drop is the same across
all the branches of a set of resistors in parallel, we know that the voltage drop across both
resistors will be 12 V. That means we just need to apply Ohm’s Law twice, once to each
resistor:
If we apply Ohm’s Law to the total resistance in the system, we find that = ( 12 V)/(4 )
= 3 A. As we might expect, the total current through the system is the sum of the current
through each of the two branches. The current is split into two parts when it branches
into the resistors in parallel, but the total current remains the same throughout the whole
circuit. This fact is captured in the junction rule we will examine when we look at
Kirchhoff’s Rules.
WHAT IS THE POWER DISSIPATED IN THE RESISTORS?
Recalling that P = I^2 R, we can solve for the power dissipated through each resistor
individually, and in the circuit as a whole. Let be the power dissipated in , the
power dissipated in , and the power dissipated in.
Note that + =.
Circuits with Resistors in Parallel and in Series
Now that you know how to deal with resistors in parallel and resistors in series, you have
all the tools to approach a circuit that has resistors both in parallel and in series. Let’s
take a look at an example of such a circuit, and follow two important steps to determine
the total resistance of the circuit.