The conducting block of square cross-section (3mm by 3mm) of length 25cm is connected at the end by a block of the same cross-section but of the length of 40cm. first block made of material with resistivity of 1.5 x 10^-6ωm and second block’s resistivity is 1.8 × 10^-6ωm. find the resistance of the entire block between two ends.

RT=R1+R2=0.122Ω

Explanation: In order to explain this problem we have to consider the definition of resistence, which is given by:

R=ρ*L/A  where ρ is the resistivity. L and A are the length and the area of the conductor.

The resistence for block 1 is given by:

R1=ρ1*L1/A =(1.5*10^-6*0.25)/9*10^-6=0.042 Ω

Then the resistance for the second conducting block

R2=ρ2*L2/A=(1.8*10^-6*0.40)/9*10^-6=0.08 Ω

Finally as the second block is connected in serie with block 1 ( the second block is connected at the end of the first block), the total resistence is equal R1+R2, then

RT=R1+R2=(0.042+0.8) Ω=0.122 Ω

1 a

2 b

3 b

4 d

5 d

6 d

explanation:

an energy transformation flow diagram is shown

where is it ?