a 20.0 ohm and 60.0 ohm resistor are connected in series to a 9.00 v battery. what is the voltage drop…

a 20.0 ohm and 60.0 ohm resistor are connected in series to a 9.00 v battery. what is the voltage drop across the 20.0 ohm resistor? (hint: how much current flows across it?) (unit = v)

a 20.0 ohm and 60.0 ohm resistor are connected in series to a 9.00 v battery. what is the voltage drop across the 20.0 ohm resistor? (hint: how much current flows across it?) (unit = v)

Answer

Explanation:

Step1: Calculate total resistance

In series, $R_{total}=R_1 + R_2$. So $R_{total}=20.0\Omega+60.0\Omega = 80.0\Omega$.

Step2: Calculate the current in the circuit

Using Ohm's law $I=\frac{V}{R}$, where $V = 9.00V$ and $R = R_{total}=80.0\Omega$. So $I=\frac{9.00V}{80.0\Omega}=0.1125A$.

Step3: Calculate voltage drop across 20 - ohm resistor

Using Ohm's law again for the 20 - ohm resistor, $V_1=IR_1$. Substitute $I = 0.1125A$ and $R_1 = 20.0\Omega$. So $V_1=0.1125A\times20.0\Omega = 2.25V$.

Answer:

$2.25V$