If with steady-state heat flow established, you double the thickness of a wall built from solid uniform material, the rate of heat loss for a given temperature difference across the thickness will:

a) become 1/ 2 of its original value.
b) become one-half of its original value.
c) become four times its original value.
d) also double.
e) become one-fourth of its original value.

A) Become 1/ 2 of its original value.

B) Become one-half of its original value.

Explanation:

As we know that heat transfer trough the wall of thickness t given as

Where

K=Thermal conductivity of the wall

A= Cross sectional area of the wall

t=Thickness of the wall

ΔT=temperature difference across the wall surfaces.

When the thickness become double ,lets sat t' = 2 t

Then new heat transfer

Therefore the new heat transfer become half of the original.

electrical power

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete