ok here we go ... its a series-parellel circuit with 5 resistors.

12 volt battery- wire goes up then one wire branches up at 45 deg with a 10 ohm resistor then down at a 45 deg with a 10 ohm resistor, a second wire branches down at 45 deg with a 20 ohm resistor then up at a 45 deg with a 25 ohm resistor, now both wires join up again as one (kinda looks like a leaf) and into a 50 ohm resistor and back to the battery
Looking for how to solve this one to find the current for each, the voltage for each and the power consumed by each..

If I am reading it correctly, you have a parallel circuit in which two series branches are in parallel.

The resistance of the series branches are 20 ohm(10+10), and 45 ohm(20+25). Those in parallel offer a total resistance of

Rp=20*45/(20+45) figure that out.

Then, that is in series with the 50 ohm resitor.

Find total current. Find voltage across the 50 ohm resistor (Itotal*50).

Now you know voltage across the parallel branch, so you can easily find current in each branch.

Hi Bob, just to make sure... I figure out the top and bottom .. and not the left and right .. is that correct?

To solve this series-parallel circuit, we can break it down into smaller, easier-to-analyze parts. Let's start by identifying the resistors and their configurations.

In this circuit, we have five resistors: R1 = 10 ohms (first branch up), R2 = 10 ohms (second branch down), R3 = 20 ohms (second branch up), R4 = 25 ohms (first branch down), and R5 = 50 ohms (connecting both branches).

1. **Current (I)**: To find the current flowing through each resistor, we need to calculate the total current (I_total) first. Based on Ohm's Law, I = V / R, where V is the voltage and R is the resistance.

Given that we have a 12-volt battery, we can calculate the total current by summing up the individual currents in each branch. Each branch can be analyzed separately as series circuits, and then combine them as a parallel circuit.

First, let's calculate the current in the first branch (R1 and R4):
- Branch 1:
- Voltage drop (V1) = 12 volts
- Resistance (R1 + R4) = 10 + 25 = 35 ohms
- Current in branch 1 (I1) = V1 / (R1 + R4) = 12 V / 35 Ω

Now, let's calculate the current in the second branch (R2, R3, and R5):
- Branch 2:
- Voltage drop (V2) = 12 volts
- Resistance (R2 + R3 + R5) = 10 + 20 + 50 = 80 ohms
- Current in branch 2 (I2) = V2 / (R2 + R3 + R5) = 12 V / 80 Ω

Finally, summing up the individual currents, we get the total current (I_total) = I1 + I2.

2. **Voltage (V)**: The voltage across each resistor can be found using Ohm's Law (V = I * R), where I is the current flowing through the resistor and R is its resistance.

For the first branch:
- Voltage across R1 = I1 * R1
- Voltage across R4 = I1 * R4

For the second branch:
- Voltage across R2 = I2 * R2
- Voltage across R3 = I2 * R3
- Voltage across R5 = I2 * R5

3. **Power (P)**: The power consumed by each resistor can be calculated using the formula P = I * V.

For the first branch:
- Power consumed by R1 = I1 * V1 (or I1^2 * R1)
- Power consumed by R4 = I1 * V4 (or I1^2 * R4)

For the second branch:
- Power consumed by R2 = I2 * V2 (or I2^2 * R2)
- Power consumed by R3 = I2 * V3 (or I2^2 * R3)
- Power consumed by R5 = I2 * V5 (or I2^2 * R5)

By following these steps, you can find the current, voltage, and power for each resistor in the series-parallel circuit you described.