Question

Three resistors of 1.00 × 102 Ω, 3.90 × 103 Ω, and 1.00 × 103 Ω are connected in series across a 2.00 × 102 V battery. What is the voltage drop across the resistor of value 1.00 × 103 Ω?

Option 1: 19.0 V
Option 2: 40.0 V
Option 3: 60.0 V
Option 4: 156 V

Answer 1

The voltage drop across the 1.00 × 103 Ω resistor in the series circuit is 40.0 V, which is determined using the total resistance of the circuit and Ohm's Law.

Step-by-step explanation:

To find the voltage drop across a resistor in a series circuit, it's important to know the total resistance and the current flowing through the circuit. The total resistance (Rtotal) is the sum of all three resistors, which is 1.00 × 102 Ω + 3.90 × 103 Ω + 1.00 × 103 Ω = 5.00 × 103 Ω. Using Ohm's Law, the total current (I) for the circuit with a 2.00 × 102 V battery can be calculated as I = V / Rtotal = 2.00 × 102 V / 5.00 × 103 Ω = 0.040 A. The voltage drop across the resistor of value 1.00 × 103 Ω is then V = I × R = 0.040 A × 1.00 × 103 Ω = 40 V. Therefore, the correct option is 40.0 V.