Question

Use the circuit diagram to determine the voltage drop across the 8 Ω resistor.

Answer 1

First we need to find the current flowing in the circuit. The three resistors are in series, so the equivalent resistance of the circuit is the sum of the three resistances:
$R_{eq} = R_1 + R_2 + R_3 = 10.0 \Omega + 8.0 \Omega +6.0 \Omega = 24.0 \Omega$

Then we can apply Ohm's law to the whole circuit, to find the current flowing:
$I= \frac{V}{R_{eq}}= \frac{12 V}{24.0 \Omega}=0.5 A$

And now we can apply Ohm's law to the resistor of $8.0 \Omega$ to find the voltage drop across it:
$V_{8\Omega} =IR=(0.5 A)(8.0 \Omega)=4 V$

Answer 2

Answer:

4 V

Explanation: