Electrical circuit with a 5V power supply and multiple resistors in series and parallel.

Let's solve the circuit shown in your image step by step.

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Circuit Description:

We have a 5 V battery connected to a network of resistors:

- R1 = 4 Ω
- R2 = 4 Ω
- R3 = 8 Ω
- R4 = 10 Ω
- R5 = 4 Ω
- R6 = 2 Ω
- R7 = 2 Ω

The goal is likely to find:
- Total resistance of the circuit
- Total current from the battery
- Currents and voltages across individual resistors (if needed)

But since no specific question was asked, I'll compute the equivalent resistance, total current, and then break it down to find currents through key branches.

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Step 1: Identify Series and Parallel Combinations

Let’s go from right to left.

#### Rightmost branch: R6 and R7
R6 = 2 Ω and R7 = 2 Ω are in series.
So their combined resistance:
\[
R_{67} = R6 + R7 = 2 + 2 = 4\ \Omega
\]

Now this 4 Ω is in parallel with R5 = 4 Ω.

So, R5 || R67:
\[
R_{5||67} = \frac{R5 \cdot R_{67}}{R5 + R_{67}} = \frac{4 \cdot 4}{4 + 4} = \frac{16}{8} = 2\ \Omega
\]

This combination (R5 || R67) is in series with R4 = 10 Ω.

So:
\[
R_{4+5||67} = 10 + 2 = 12\ \Omega
\]

Now look at the middle branch: R2 and R3 in series.

R2 = 4 Ω, R3 = 8 Ω → in series:
\[
R_{23} = 4 + 8 = 12\ \Omega
\]

Now we have two branches in parallel:
- One is R23 = 12 Ω
- The other is R4 + (R5 || R67) = 12 Ω

So both branches are 12 Ω in parallel.

Their equivalent resistance:
\[
R_{\text{parallel}} = \frac{12 \cdot 12}{12 + 12} = \frac{144}{24} = 6\ \Omega
\]

Now, this 6 Ω is in series with R1 = 4 Ω.

So total resistance:
\[
R_{\text{total}} = R1 + R_{\text{parallel}} = 4 + 6 = 10\ \Omega
\]

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Step 2: Total Current from Battery

Voltage = 5 V
Total Resistance = 10 Ω

Using Ohm’s Law:
\[
I_{\text{total}} = \frac{V}{R_{\text{total}}} = \frac{5}{10} = 0.5\ \text{A}
\]

So, the current leaving the battery is 0.5 A.

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Step 3: Voltage Drops

Current through R1 = 0.5 A
R1 = 4 Ω → Voltage drop:
\[
V_{R1} = I \cdot R1 = 0.5 \cdot 4 = 2\ \text{V}
\]

So voltage remaining for the rest of the circuit:
\[
V_{\text{parallel}} = 5 - 2 = 3\ \text{V}
\]

This 3 V appears across both parallel branches:
- Branch 1: R2 + R3 (12 Ω)
- Branch 2: R4 + (R5 || R67) (12 Ω)

Since both branches have equal resistance (12 Ω), they split the current equally.

So current in each branch:
\[
I_{\text{branch}} = \frac{3\ \text{V}}{12\ \Omega} = 0.25\ \text{A}
\]

So:
- Current through R2 and R3: 0.25 A
- Current through R4 and the right-hand side: 0.25 A

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Step 4: Break Down Right Side

Current through R4 = 0.25 A
R4 = 10 Ω → Voltage drop:
\[
V_{R4} = 0.25 \cdot 10 = 2.5\ \text{V}
\]

Remaining voltage for the parallel part (R5 || R67) is:
\[
V_{\text{right}} = 3 - 2.5 = 0.5\ \text{V}
\]

Now, R5 = 4 Ω and R67 = 4 Ω are in parallel, both have same resistance → equal current.

So current through R5:
\[
I_{R5} = \frac{0.5\ \text{V}}{4\ \Omega} = 0.125\ \text{A}
\]

Current through R67 = 0.125 A

But R6 and R7 are in series (both 2 Ω), so same current:
\[
I_{R6} = I_{R7} = 0.125\ \text{A}
\]

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Step 5: Summary of Results

| Resistor | Resistance | Current (A) | Voltage (V) |
|---------|------------|-------------|-------------|
| R1 | 4 Ω | 0.5 | 2.0 |
| R2 | 4 Ω | 0.25 | 1.0 |
| R3 | 8 Ω | 0.25 | 2.0 |
| R4 | 10 Ω | 0.25 | 2.5 |
| R5 | 4 Ω | 0.125 | 0.5 |
| R6 | 2 Ω | 0.125 | 0.25 |
| R7 | 2 Ω | 0.125 | 0.25 |

> Note: Voltages are calculated as \( V = I \cdot R \)

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Final Answers:

- Total Resistance: 10 Ω
- Total Current: 0.5 A
- Voltage across R1: 2 V
- Voltage across the parallel branches: 3 V
- Current in R2-R3 branch: 0.25 A
- Current in R4-R5-R6-R7 branch: 0.25 A
- Current in R5: 0.125 A
- Current in R6 and R7: 0.125 A each

Let me know if you want power dissipation or something else!