Question

A Series ˚uit has two 8Ω loads with 6V drops across each one. If a third 8Ω load is added to the

˚uit, what will be the voltage drop across the first load in the ˚uit?
A. 8V
B. 6V
C. 4V
D. 2V

Answer 1

After adding a third 8Ω load to a series circuit with two existing 8Ω loads, each having a 6V drop, the voltage drop across the first load becomes 4V due to the change in total resistance and the current adjusting to maintain the total voltage.

Step-by-step explanation:

Given that a series circuit initially has two 8Ω loads each with a 6V drop, the total voltage in the circuit is the sum of the voltage drops across each load, which is 12V (2 loads × 6V). If a third load of 8Ω is added, the total resistance of the circuit becomes 24Ω (3 loads × 8Ω). To maintain the same total voltage of 12V, Ohm's law (V=IR) dictates that the circuit's current will adjust to maintain the total voltage. Thus, with 12V and 24Ω, the current in the circuit would be 0.5A (12V ÷ 24Ω). The voltage drop across each 8Ω load is then found by multiplying the current by the resistance of one load (V=IR), leading to a 4V drop across each load (0.5A × 8Ω). Therefore, the voltage drop across the first load in the circuit after adding the third load would be 4V.