Dimensional Analysis and units of voltage

[Nusc]

Homework Statement

The first equation below has units of voltage.

Homework Equations

[tex]
V(t)=IR(1-e^{t/RC}) \rightarrow ln|IR - V(t)| = ln|IR|-t/RC
[/tex]

The Attempt at a Solution

the logarithm has units of voltage, but what does ln|IR| and ln|V| physically mean?

 

[Nusc]

The purpose of this question is that if you plot the natural logarithm of the voltage versus time, it should yield a straight line with slope -1/RC and intercept ln|V_c|.

From which you can determine the capacitance given a known resistance. Or maybe I should post this in the general physics forum?

 

[nlightner]

Dimensional analysis is a powerful tool used in science to check the validity of equations and to ensure that the units on both sides of the equation are consistent. In the context of voltage, it is important to understand that voltage is a measure of electric potential difference, which is the difference in electric potential energy per unit charge between two points. Therefore, the units of voltage must reflect this relationship.

In the given equation, V(t) is the voltage at a given time t, and it is equal to the product of current (I) and resistance (R), multiplied by a term that accounts for the charging or discharging of a capacitor over time. This term, 1-e^(t/RC), is dimensionless and does not have units.

The logarithm functions, ln|IR| and ln|V|, are used to solve for the voltage at a specific time t. They do not have units of voltage, but rather they represent the natural logarithm of a ratio of voltage (V) to the product of current (I) and resistance (R). This can be interpreted as the ratio of the electric potential energy (represented by voltage) to the amount of work done (represented by the product of current and resistance) to produce that energy.

In summary, the units of voltage in this equation are consistent, as V(t) has units of voltage, and ln|IR| and ln|V| do not have units of voltage but are used to solve for V(t) at a specific time t.