The equation \( \mathbf{E} = \rho \mathbf{J} \) is known as the electrical conductivity equation. Here:
From this equation, we can express the electric field in terms of current density:
\[ \mathbf{E} = \rho \mathbf{J} \]
Now, consider **Ohm's law**, which states that the current density \( \mathbf{J} \) is proportional to the electric field \( \mathbf{E} \) and the material's conductivity \( \sigma \) (the inverse of resistivity). So, we can write:
\[ \mathbf{J} = \sigma \mathbf{E} \]
Since \( \sigma = \frac{1}{\rho} \), we can substitute this into the above equation:
\[ \mathbf{J} = \frac{1}{\rho} \mathbf{E} \]
Rearranging the equation, we get:
\[ \mathbf{E} = \rho \mathbf{J} \]
This is exactly the form of the equation we started with, so we have derived Ohm's law from the equation \( \mathbf{E} = \rho \mathbf{J} \).
Ohm's law assumes that the material has a constant resistivity \( \rho \) and that the current is proportional to the applied voltage (i.e., linear response). However, there are conditions under which Ohm's law does not hold:
Thus, Ohm’s law is not valid in situations where the material’s resistivity is not constant or when extreme conditions like high electric fields or temperatures cause a non-linear relationship between voltage and current.
A battery of emf \( E \) and internal resistance \( r \) is connected to a rheostat. When a current of 2A is drawn from the battery, the potential difference across the rheostat is 5V. The potential difference becomes 4V when a current of 4A is drawn from the battery. Calculate the value of \( E \) and \( r \).
परसेवा का आनंद — 120 शब्दों में रचनात्मक लेख लिखिए:
Answer the following questions:
[(i)] Explain the structure of a mature embryo sac of a typical flowering plant.
[(ii)] How is triple fusion achieved in these plants?
OR
[(i)] Describe the changes in the ovary and the uterus as induced by the changes in the level of pituitary and ovarian hormones during menstrual cycle in a human female.