Current density \( \mathbf{j} \) is the current per unit area and is a vector. It is given by:
\[ \mathbf{j} = \frac{I}{A} \]
The drift velocity \( v_d \) is related to the electric field \( E \) as:
\[ v_d = -\frac{eE\tau}{m} \]
Where \( e \) is the charge of an electron, \( \tau \) is the relaxation time, and \( m \) is the mass of the electron.
The current density is given by the product of the number of charge carriers \( n \), the charge of the carriers \( e \), and their drift velocity \( v_d \):
\[ \mathbf{j} = n(-e)v_d = n \cdot (-e) \cdot \left( -\frac{eE\tau}{m} \right) \]
Simplifying this, we get:
\[ \mathbf{j} = \frac{ne^2\tau E}{m} \]
Thus, the conductivity \( \alpha \) is:
\[ \alpha = \frac{ne^2\tau}{m} \]
The Wheatstone bridge is used to measure an unknown resistance. It is balanced when the following condition is met:
\[ \frac{R_1}{R_2} = \frac{R_3}{R_4} \]
In this case, there is no current flowing through the galvanometer, and the bridge is considered balanced.

Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.
Current passing through a wire as function of time is given as $I(t)=0.02 \mathrm{t}+0.01 \mathrm{~A}$. The charge that will flow through the wire from $t=1 \mathrm{~s}$ to $\mathrm{t}=2 \mathrm{~s}$ is: