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 principle of dimensional consistency.
In any physically meaningful equation: LHS and RHS must have the same dimensions Also, the argument of an exponential function must always be dimensionless.

Step 2: Check option (1).
Given, \[ I=I_0\left(e^{\frac{2V}{V_0}}+1\right) \] Since \[ \frac{V}{V_0} \] is dimensionless, the exponential term is valid.
Also, \[ e^{\frac{2V}{V_0}}+1 \] is dimensionless.
Therefore, RHS has dimensions: \[ [I_0] \] which is the dimension of current.
Hence, option (1) is dimensionally correct.

Step 3: Check option (2).
Given, \[ I=I_0\left(e^{\frac{V}{2V_0}}-1\right) \] Again, \[ \frac{V}{V_0} \] is dimensionless.
Thus, \[ e^{\frac{V}{2V_0}}-1 \] is dimensionless.
Hence, RHS has dimensions: \[ [I_0] \] which matches current.
So, option (2) is dimensionally correct.

Step 4: Check option (3).
Given, \[ I=I_0V_0\left(e^{\frac{V}{V_0}}-1\right) \] The exponential term is dimensionless.
Therefore, RHS has dimensions: \[ [I_0][V_0] \] That is, \[ (\text{current})(\text{potential}) \] But LHS has dimensions only of current.
Thus, dimensions do not match.
Hence, option (3) is dimensionally incorrect.

Step 5: Check option (4).
Given, \[ I=I_0\left(\frac{V}{V_0}\right)\left(e^{\frac{V}{V_0}}-1\right) \] Both \[ \frac{V}{V_0} \] and \[ e^{\frac{V}{V_0}}-1 \] are dimensionless.
Therefore, RHS has dimensions: \[ [I_0] \] which matches current.
So, option (4) is dimensionally correct.

Step 6: Final conclusion.
Hence, the dimensionally incorrect expression is \[ \boxed{(3)} \]