How do you find the current and voltage in the circuit shown on the attached image?

This problem involves the current and voltage distribution in parallel and series circuits, as well as the equivalent resistance of these circuits.

Notice that in the circuit on the attached image, the 30 ` ` and `6 Omega` resistors are connected in parallel, and the `15 Omega` resistor is connected to them in series. We can calculate the equivalent resistance of this network:

The equivalent resistance of the resistors `R_1` and `R_2` is determined by the formula

`1/R = 1/R_1 + 1/R_2`

so the equivalent resistance of `30 Omega` and `6 Omega` resistors will be the reciprocal of

`1/30 + 1/6 = 6/30 = 1/5` , or `5 Omega` .

Since the `15 Omega` resistor is connected in series, the equivalent resistance of the whole circuit will be the sum of `15 Omega` and `5 Omega` : 15 + 5 = 20 `Omega` .

Now that we know the resistance of the circuit, we can calculate the current in this circuit, supplied by the 120 V battery:

`I = 120/20 = 6 A` .

This current I will flow through the battery, as well as through the 15 `Omega` resistor. The voltage drop on this resistor is determined by the Ohm's Law:

`V = 6A * 15 Omega = 90 V` .

The voltage drop on the `30 Omega` and  `6 Omega` resistors is the same (because they are connected in parallel) and equal to the remaining voltage: 120 V - 90 V = 30V.

The total current I = 6 A splits among these two resistors in a way such that the current is inversely proportional to the resistance (this, again, follows from the Ohm's Law and the fact that the voltage is the same.) So, since one resistance is 5 times greater than the other, the current through `30 Omega` resistor is 5 times less than the current through `6 Omega` resistor. Since the total current is 6 A, the former current is 1 A and the latter current is 5 A.

To summarize,

for `15 Omega ` resistor, I = 6A and V = 90V;

for `30 Omega` resistor, I = 1A and V = 30V;

for `6 Omega` resistor, I = 5A and V = 30V.