Let the resistance of the electric lamp be R_{lamp}.

Current (I) = 1 A

Resistance of conductor (R_{conductor}) = 5Ω

Potential difference of battery (V) = 10 V

Since the lamp and the conductor is connected in series, thus same current 1 A will pass through both of them.

Using Ohm's law,

R_{net }= \(\frac VI\)

R_{net }= \(\frac{10}1\)

R_{net} = 10Ω

R_{net }= R_{lamp} + R_{conductor}

⇒ 10 = R_{lamp} + 5

⇒ R_{lamp} = 5Ω

Potential difference across lamp,

V_{lamp} = I × R_{lamp} = 1 × 5 = 5V

When 10Ω resistor is connected parallel to the series combination of lamp and conductor (R_{net} = 5 + 5 = 10Ω) then the equivalent resistance,

\(\frac 1{R_{eq}} = \frac 1{10} + \frac 1{10} \)

\(= \frac 2{10}\)

\(= \frac 15\)

⇒ R_{eq} = 5Ω

Using Ohm's law,

\(I' = \frac V{R_{eq}}\)

⇒ \(I' = \frac{10}5\)

⇒ \(I' = 2A\)

Current will distribute equally in two parallel parts.

Thus,\(\frac{I'}2 = 1A\) current will pass through both the lamp and the resistor of 5Ω (because they are connected in series).

Potential difference across the lamp (R_{lamp}) = 5Ω

V'_{lamp }= 1 × 5 = 5V

Hence, there will be no change in current through the conductor of resistance 5Ω, and potential difference across the lamp.