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Difficulty: EasySeries Circuits

Three resistors with values of 10 Ω10\ \Omega, 15 Ω15\ \Omega, and 25 Ω25\ \Omega are connected in series across a direct current (DC) voltage source. What is the total equivalent resistance of the circuit?

  1. A
    4.8 Ω4.8\ \Omega
  2. 50 Ω50\ \OmegaAnswer
  3. C
    25 Ω25\ \Omega
  4. D
    50 kΩ50\text{ k}\Omega

Answer

The total equivalent resistance is 50 Ω50\ \Omega.
In a series circuit, electrical current flows through each component sequentially along a single path. Therefore, the total opposition to current flow is the sum of the individual resistor values: RT=10 Ω+15 Ω+25 Ω=50 ΩR_T = 10\ \Omega + 15\ \Omega + 25\ \Omega = 50\ \Omega.

Step-by-Step Solution

1
Identify the circuit configuration and the formula for total series resistance.
RT=R1+R2+R3R_T = R_1 + R_2 + R_3
In a series circuit, there is only one continuous path for electric current, meaning the opposition to current flow is cumulative.
2
Substitute the given resistor values into the series formula.
RT=10 Ω+15 Ω+25 ΩR_T = 10\ \Omega + 15\ \Omega + 25\ \Omega
Each component contributes its full resistive value directly to the total opposition.
3
Calculate the arithmetic sum.
RT=50 ΩR_T = 50\ \Omega
Simple addition yields the equivalent circuit resistance.

Key Concept

In a series circuit, total equivalent resistance (RTR_T) equals the direct sum of all individual resistances (RT=R1+R2+⋯+RnR_T = R_1 + R_2 + \dots + R_n).
Estimated Time:45s
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Topics

Series Circuits

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