Three resistors with resistance 200Ω, 500Ω and 1kΩ are connected in series. A 6v battery is connected to…

1. Analyze the Question

We are given three resistors connected in series: $R_1 = 200\,\Omega$, $R_2 = 500\,\Omega$, and $R_3 = 1000\,\Omega$ (since $1\text{k}\Omega = 1000\,\Omega$). They are connected to a $6\text{V}$ battery. We need to find the potential difference (voltage drop) across the $200\,\Omega$ resistor.

2. Explain the Concept

• Series Circuit: In a series circuit, the current ($I$) flowing through each resistor is the same.
• Total Resistance ($R_T$): For resistors in series, $R_T = R_1 + R_2 + R_3$.
• Ohm's Law: $V = IR$, where $V$ is voltage, $I$ is current, and $R$ is resistance.
• Voltage Divider Rule: The voltage across a specific resistor in a series circuit is proportional to its resistance relative to the total resistance: $V_x = V_{total} \times \frac{R_x}{R_T}$.

3. Show the Solution

Step 1: Calculate the total resistance ($R_T$)

Step 2: Calculate the total current ($I$) in the circuit Using Ohm's Law ($I = \frac{V}{R_T}$):

Step 3: Calculate the potential difference across the $200\,\Omega$ resistor ($V_{200}$) Using Ohm's Law ($V = IR$):

Rounding to two decimal places, we get 0.71V.

4. Identify the Answer

Comparing our result to the given options, the value matches option A.

5. Exam Tip

When dealing with series circuits, always remember that the largest resistor will always have the largest voltage drop across it. If you ever get a voltage drop that is higher than the battery's total voltage, you know you've made a calculation error!