Question

The expressions below give current \(I\) through an electronic component as a function of applied potential \(V\). \(I_0\) and \(V_0\) are constants having dimensions of current and potential respectively. Which of the following are dimensionally incorrect?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

The argument of exponential, logarithmic, trigonometric, and inverse trigonometric functions must always be dimensionless.

Answer 1

The Correct Option is C

Step 1: Recall the fundamental rule of dimensional consistency.
Any physically valid equation must have the same dimensions on both its left-hand side and right-hand side. Additionally, the argument of a mathematical function such as an exponential must always be dimensionless. If either condition is violated, the expression is dimensionally incorrect.
Step 2: Identify the dimensions of the given constants.
We are told that $I_0$ has dimensions of current, i.e., $[I_0] = A$ (ampere), and $V_0$ has dimensions of potential, i.e., $[V_0] = V$ (volt). The quantity $V/V_0$ is dimensionless since both $V$ and $V_0$ have dimensions of potential, and their ratio cancels.
Step 3: Check options (1) and (2) for dimensional correctness.
Both option (1), $I = I_0(e^{2V/V_0} + 1)$, and option (2), $I = I_0(e^{V/2V_0} - 1)$, follow the form $I = I_0 \times (\text{dimensionless factor})$. Since the exponential arguments $2V/V_0$ and $V/2V_0$ are dimensionless, the exponential results are dimensionless. Multiplying by $I_0$ gives dimensions of current, which matches the left-hand side. So options (1) and (2) are dimensionally correct.
Step 4: Analyse option (3) for dimensional correctness.
Option (3) has the form $I = I_0 V_0 (e^{V/V_0} - 1)$. The exponential factor is dimensionless, so the right-hand side has dimensions $[I_0][V_0] = A \cdot V = \text{watt}$. This is not equal to the dimensions of current ($A$). Therefore, option (3) is dimensionally incorrect.
Step 5: Understand why option (3) fails physically.
The product $I_0 V_0$ has units of power (watts), not current. No matter what dimensionless factor multiplies it, the result will never have units of current. This is the fundamental error in option (3): an extra factor of $V_0$ has been introduced without a compensating inverse factor.
Step 6: Conclude which option is dimensionally incorrect.
Option (3) is the only expression among the given choices that fails dimensional consistency, because its right-hand side has dimensions of power rather than current. \[ \boxed{\text{Option (3) is dimensionally incorrect}} \]