In basic circuit analysis, which equation correctly gives the voltage drop across an individual resistor carrying current I with resistance R?

Difficulty: Easy

Correct Answer: I x R

Explanation:


Introduction / Context:
Ohm's law connects voltage, current, and resistance, allowing you to compute voltage drops across components. Correctly recalling the form of this relationship is crucial for sizing parts, predicting behavior, and diagnosing faults in DC and AC circuits alike.


Given Data / Assumptions:

  • Linear resistor with resistance R (ohms).
  • Electrical current I (amperes) flows through the resistor.
  • We want the voltage drop across that resistor.


Concept / Approach:
Ohm's law states V = I * R for an ohmic element. This is the constitutive relation for resistors and follows from the proportionality between electric field and current density in a uniform conductor. Power relations such as P = I^2 * R or P = V * I are derived, not primary for finding voltage drop directly.


Step-by-Step Solution:
Start with Ohm's law: V = I * R.Identify given quantities: I (current), R (resistance).Compute voltage drop using multiplication of I and R.Therefore, the correct expression is I x R.


Verification / Alternative check:
If power P is known, P = I^2 * R and P = V * I. Solving P = V * I for V gives V = P / I. Substituting P = I^2 * R yields V = (I^2 * R) / I = I * R, confirming the same relationship from power formulas.


Why Other Options Are Wrong:

  • V x R: Dimensional mismatch; multiplying voltage by resistance does not yield voltage.
  • I2 x R: That is the power dissipated (I^2 * R), not the voltage drop.
  • V x I: Also a power expression (P = V * I), not voltage.


Common Pitfalls:

  • Confusing V = I * R with power formulas, especially under exam pressure.
  • For AC with complex impedances, use magnitudes and phases: V = I * Z. For a pure resistor, Z = R.


Final Answer:
I x R

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