Question

question 8 of 25 a series circuit is shown in the diagram below. what is the potential drop across r₂? (r₁ = 20 ω, r₂ = 40 ω, r₃ = 60 ω, v = 60 v) (ohms law: v = ir)

Explanation:

Step1: Calculate total resistance

In a series - circuit, $R_{total}=R_1 + R_2+R_3$. Given $R_1 = 20\Omega$, $R_2 = 40\Omega$, $R_3 = 60\Omega$, so $R_{total}=20 + 40+60=120\Omega$.

Step2: Calculate the current in the circuit

Using Ohm's law $V = IR$, we can find the current $I$. Given $V = 60V$ and $R_{total}=120\Omega$, then $I=\frac{V}{R_{total}}=\frac{60}{120}=0.5A$.

Step3: Calculate the potential drop across $R_2$

Using Ohm's law again for $R_2$, $V_2=IR_2$. Since $I = 0.5A$ and $R_2 = 40\Omega$, then $V_2=0.5\times40 = 20V$.

Answer:

A. 20 V