91影视

The emf of the battery,$V_A-V_D=3.5V$

The electric field thick wire us very small due to this the potential difference across it is negligible.

The voltage difference,$V_B-V_{A}0V$

The voltage difference,$V_D-V_C=0V$

The average drift speed of the mobile electron is given by the product of the electron mobility and electric field.

$\mathit{V}\mathbf{=}\mathit{U}\mathit{E}$ 鈥︹�� (i)

Here, U is the electron mobility and E is the electric field.

The expression to calculate the current in the circuit is given as follows.

$\mathit{i}\mathbf{=}\mathit{n}\mathit{A}\mathit{v}$ 鈥︹�� (ii)

Here, A is the area of the wire.

According to the Kirchhoff鈥檚 Voltage law, the algebraic sum of the voltage in the circuit is always equal to zero.

$$\begin{gathered} 鈭�V=0 \\ (V_B-V_A)+(V_C-V_B)+(V_D-V_C)+(V_A-V_D)=0 \\ 0+(V_C-V_{\mathbf{B}})+0+3.5=0 \\ (V_C-V_B)=-3.5V \end{gathered}$$

The electrical field in the circuit is given by,

$E=\frac{鈭�V}{L}$

Derive the expression for the current in the circuit.

Substitute$uE$ for V into equation (ii).

$i=nAuE$

Substitute$鈭�V/L$ for$E$ into above equation.

$i=nAu\frac{鈭�V}{L}$

From the above equation, it is clear that, to calculate the current in the circuit we need property of the resistor. Therefore, the required information to calculate the current in the circuit is not enough.

Hence, the potential difference $V_C-V_B$is $3.5V$and the information is not enough to calculate the current in the circuit.