Series circuit with a 9V battery and three resistors (R1 = 3 kΩ, R2 = 10 kΩ, R3 = 5 kΩ) illustrating current flow.

Let's solve the problem step by step based on the circuit shown in the image.

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Given:

- Voltage source: $ V = 9 \, \text{V} $
- Resistors:
- $ R_1 = 3\, \text{k}\Omega $
- $ R_2 = 10\, \text{k}\Omega $
- $ R_3 = 5\, \text{k}\Omega $

The resistors are connected in a series configuration around a single loop (a simple series circuit), as there is only one path for current to flow.

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Step 1: Total Resistance

Since all resistors are in series, total resistance $ R_{\text{total}} $ is:

$$
R_{\text{total}} = R_1 + R_2 + R_3
$$

$$
R_{\text{total}} = 3\, \text{k}\Omega + 10\, \text{k}\Omega + 5\, \text{k}\Omega = 18\, \text{k}\Omega
$$

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Step 2: Current in the Circuit

Using Ohm’s Law:

$$
I = \frac{V}{R_{\text{total}}} = \frac{9\, \text{V}}{18\, \text{k}\Omega} = 0.5\, \text{mA}
$$

So, the current $ I = 0.5\, \text{mA} $ flows through the entire circuit (same through all resistors in series).

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Step 3: Voltage Across Each Resistor

We can calculate the voltage drop across each resistor using $ V = I \times R $:

- $ V_1 = I \times R_1 = 0.5\, \text{mA} \times 3\, \text{k}\Omega = 1.5\, \text{V} $
- $ V_2 = I \times R_2 = 0.5\, \text{mA} \times 10\, \text{k}\Omega = 5\, \text{V} $
- $ V_3 = I \times R_3 = 0.5\, \text{mA} \times 5\, \text{k}\Omega = 2.5\, \text{V} $

Check: $ V_1 + V_2 + V_3 = 1.5 + 5 + 2.5 = 9\, \text{V} $ ✔

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Final Answers:

- Total current: $ I = 0.5\, \text{mA} $
- Voltage drops:
- $ V_{R1} = 1.5\, \text{V} $
- $ V_{R2} = 5\, \text{V} $
- $ V_{R3} = 2.5\, \text{V} $

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Explanation:

This is a simple series circuit with a 9 V battery and three resistors in series. In series circuits:
- The current is the same everywhere.
- The voltages add up to the supply voltage.
- The total resistance is the sum of individual resistances.

The direction of current is clockwise (as indicated by the arrow), starting from the positive terminal of the battery, going through $ R_1 $, $ R_2 $, $ R_3 $, and back to the negative terminal.

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Let me know if you'd like to find power dissipated or analyze node voltages!