91Ó°ÊÓ

When current is flowing through a conducting wire then due to the resistance of the wire, energy in the form of heat is dissipated.

The heat energy dissipated by a conducting wire is lower if the resistance of the conducting wire is lower.

The expression for the resistance in the connecting wire is given by,

$R=\frac{\mathrm{ÒÏ}L}{A}$

Here, $\mathrm{ÒÏ}$is the resistivity of the wire,$L$ is the length of the wire, and$A$ is the cross-sectional area of the wire.

From the expression, the resistance value of the wire will be lesser for a thicker wire per length. So, the energy dissipated in the thick connecting wires due to the resistance of the wire will also be lesser.

Due to the less resistance in the wire, the voltage drop across the copper wire is also less. Then, due to the potential difference, the electric field inside the copper wire would also be less.

Then the voltage difference or emf in the wire would be almost equal to the potential difference in the bulb.

So, the emf in the wire can be given as,

$\text{emf}≈E_{bulb}{L}_{bulb}$

Hence, the correct options are (1) Very little energy is dissipated in the thick connecting wires, and (2) The electric field in connecting wires is very small, so $\text{emf}≈E_{bulb}{L}_{bulb}$.