Take a look at this circuit and calculate the current that will run through each LED. What will the total current be? How much power is used to run the circuit below? Which LED(s) will glow the brightest?

The first step is to find the equivalent resistance of the circuit.
For the 1 kohm and 330 ohm resistors, they are in series. Thus,
equivalent resistance = 1000 + 330 = 1330 ohms
For the 330 ohm and 330 ohm resistors, they are in series
equivalent resistance = 330 + 330 = 660 ohms
Thus, 1330 ohms, 2kohms and 660 ohms are in parallel. We would calculate the equivalent resistance as shown below
1/Req = 1/1330 + 1/2000 + 1/660
1/Req = 0.002767
Req = 1/0.002767
Req = 361.398 ohms
V = IR
I = V/R
where
V is the voltage
I is the current
R is the resistance
Given that V = 9,
Total current = 9/361.398
Total current = 0.025 A
Power = IV = 0.025 x 9
Power = 0.225 W
Since the voltage passing through the circuit is the same, more current will pass through the two leds on the right since their equivalent resistance is lowest