To calculate the voltage drop ($U$) across resistor $R_1$ in this circuit, follow these steps:

Step 1: Identify the circuit type

The resistors $R_1$, $R_2$, and $R_3$ are connected in series. The total resistance of the circuit ($R_\text{total}$) is the sum of all resistances: $R_\text{total} = R_1 + R_2 + R_3$

Step 2: Calculate total resistance

Given values: $R_1 = 2 \, \Omega, \, R_2 = 4 \, \Omega, \, R_3 = 3 \, \Omega$ $R_\text{total} = 2 + 4 + 3 = 9 \, \Omega$

Step 3: Determine the total current

The total current ($I_\text{total}$) in the circuit is calculated using Ohm's Law: $I_\text{total} = \frac{V}{R_\text{total}}$ Substitute the given voltage ($V = 12 \, \text{V}$): $I_\text{total} = \frac{12}{9} = 1.33 \, \text{A}$

Step 4: Calculate the voltage drop across $R_1$

The voltage drop across $R_1$ is given by: $U_{R_1} = I_\text{total} \cdot R_1$ Substitute the values: $U_{R_1} = 1.33 \cdot 2 = 2.67 \, \text{V}$

Final Result

The voltage drop across $R_1$ is: $\boxed{2.67 \, \text{V}}$

Let me know if you need further explanations or calculations.

Questions for Further Exploration

1. How would the voltage drop change if $R_1$ increased to $5 \, \Omega$?
2. What is the voltage drop across $R_2$ and $R_3$ in this circuit?
3. How does the total current change if the supply voltage is doubled?
4. What happens to the total resistance if $R_2$ and $R_3$ are connected in parallel instead of series?
5. Can you derive the power dissipated across $R_1$?

Tip

Always double-check whether the circuit is in series or parallel, as this determines how resistances combine!