Given that the energy stored in an inductor is expressed as $U_L = \frac{1}{2}LI^2$ and the power dissipated in a resistor is $P = I^2R$, where $L$ is inductance, $R$ is resistance, and $I$ is current, determine the dimension of the ratio $\frac{L}{R}$.
$M^0 L^0 T$
Understanding the Dimensions of Inductor and Resistor Ratio
This solution explains how to determine the dimensions of the ratio $\frac{L}{R}$ using the provided formulas for energy stored in an inductor ($U_L$) and power dissipated in a resistor ($P$). We need to find the fundamental dimensions of inductance ($L$) and resistance ($R$) first.
Determining Fundamental Dimensions
Dimensions are expressions of physical quantities in terms of fundamental quantities like mass ($M$), length ($L$), time ($T$), and electric current ($A$). We use the provided formulas and the known dimensions of related quantities.
Dimensions of Energy ($U_L$)
Energy has the same dimensions as work (Force $\times$ Distance). Force dimensions are $[M L T^{-2}]$. Therefore, the dimensions of energy are: $ [U_L] = [M] \cdot [L] \cdot [L] \cdot [T^{-2}] = [M L^2 T^{-2}] $
Dimensions of Power ($P$)
Power is the rate of energy transfer, or Energy per unit time. Therefore, the dimensions of power are: $ [P] = \frac{[U_L]}{[T]} = \frac{[M L^2 T^{-2}]}{[T]} = [M L^2 T^{-3}] $
Dimensions of Current ($I$)
Electric current is considered a fundamental quantity. Its dimension is represented by $[A]$ (Ampere).
Deriving Dimensions of Inductance ($L$)
We are given the formula for energy stored in an inductor: $U_L = \frac{1}{2}LI^2$. The term $\frac{1}{2}$ is a dimensionless constant. To find the dimensions of inductance ($L$), we rearrange the formula: $ [L] = \frac{[U_L]}{[I^2]} $ Substituting the dimensions we found: $ [L] = \frac{[M L^2 T^{-2}]}{[A]^2} = [M L^2 T^{-2} A^{-2}] $
Deriving Dimensions of Resistance ($R$)
We are given the formula for power dissipated in a resistor: $P = I^2R$. To find the dimensions of resistance ($R$), we rearrange the formula: $ [R] = \frac{[P]}{[I^2]} $ Substituting the dimensions we found: $ [R] = \frac{[M L^2 T^{-3}]}{[A]^2} = [M L^2 T^{-3} A^{-2}] $
Calculating the Dimensions of the Ratio $\frac{L}{R}$
Now we need to find the dimensions of the ratio $\frac{L}{R}$ by dividing the dimensions of $L$ by the dimensions of $R$: $ \left[\frac{L}{R}\right] = \frac{[L]}{[R]} = \frac{[M L^2 T^{-2} A^{-2}]}{[M L^2 T^{-3} A^{-2}]} $ We can cancel out the dimensions that appear in both the numerator and the denominator: $[M]$, $[L^2]$, and $[A^{-2}]$. $ \left[\frac{L}{R}\right] = \frac{[T^{-2}]}{[T^{-3}]} = [T^{-2 - (-3)}] = [T^{-2 + 3}] = [T^1] = [T] $
Final Dimension Representation
The dimension of the ratio $\frac{L}{R}$ is $[T]$, which represents time. In the standard format of $[M^a L^b T^c A^d]$, this is written as: $ [M^0 L^0 T^1] $
Conclusion
The dimensions of the ratio $\frac{L}{R}$ correspond to time. Comparing this result with the given options, the correct dimension is $M^0 L^0 T$.
Important Questions from Dimensional formulae and dimensional equations
- Considering the Lorentz force $\vec{F} = q(\vec{v} \times \vec{B})$, where $F$ is force, $q$ is electric charge, and $v$ is velocity, what is the dimensional formula for magnetic flux density $B$?
The dimensions of energy are:
If force $[F]$, acceleration $[A]$ and time $[T]$ are chosen as the fundamental physical quantities. Find the dimensions of pressure.
- The characteristic impedance of free space, $Z_0$, is given by the expression $Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}}$. If $\mu_0$ represents the magnetic permeability and $\epsilon_0$ represents the electric permittivity, what are the dimensions of $Z_0$?
- Determine the dimensional formula for the quantity represented by the product of pressure and volume.
