Question

What is the voltage drop across a 2 Ω resistor connected in series with a 4 Ω resistor to a 9 V battery?

Answer 1

To calculate the voltage drop across the 2 Ω resistor in a series circuit with a 4 Ω resistor connected to a 9 V battery, we first need to find the total resistance and the current flowing through the circuit.

1. **Calculate the total resistance (R_total):**
\[
R_{\text{total}} = R_1 + R_2 = 2\, \Omega + 4\, \Omega = 6\, \Omega
\]

2. **Calculate the total current (I) using Ohm's law (V = I × R):**
\[
I = \frac{V}{R_{\text{total}}} = \frac{9\, \text{V}}{6\, \Omega} = 1.5\, \text{A}
\]

3. **Calculate the voltage drop (V_drop) across the 2 Ω resistor:**
Using Ohm's law again:
\[
V_{\text{drop}} = I \times R_1 = 1.5\, \text{A} \times 2\, \Omega = 3\, \text{V}
\]

Therefore, the voltage drop across the 2 Ω resistor is **3 V**.