three resistors of 100, 40.0, and 10.0 ohms are in series with a 24.0 v power supply. what is the voltage…

Explanation:

Step1: Calculate total resistance

For resistors in series, $R_{total}=R_1 + R_2+R_3$. So $R_{total}=100 + 40.0+10.0=150\Omega$.

Step2: Calculate the current in the circuit

Using Ohm's law $I=\frac{V}{R}$, where $V = 24.0V$ and $R = R_{total}=150\Omega$. So $I=\frac{24.0}{150}=0.16A$.

Step3: Calculate voltage drop across 40 - ohm resistor

Using Ohm's law again, $V_1 = IR_1$, with $I = 0.16A$ and $R_1=40.0\Omega$. So $V_1=0.16\times40.0 = 6.4V$.

Answer:

$6.4$