Question

The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in which the current takes different paths through a piece of material. Each of the rectangular pieces is made from the same material whose resistivity is rho = 1.50 x 10^-2 Ω·m, and the unit of length in the drawing is L0 = 5.80 cm. Each piece of material is connected to a 3.00-V battery. Find:

a. the resistance
b. the current in each case.

Answer 1

a) Ra = 0.517 Ω

Rb = 0.032 Ω

Rc = 0.129 Ω

b) Ia = 5.8A

Ib = 93.75A

Ic = 23.2 A

Step-by-step explanation:

a) The resistance is equal to:

Resistance for case a:

`R_(a) =(pL_(a) )/(A_(a) ) =(p*4*L_(0) )/(2L_(0)*L_(0) ) =(2p)/(L_(0) )`

Where

p = 1.5x10⁻²Ωm

L0 = 5.8 cm = 0.058 m

`R_(a) =(2*1.5x10^(-2) )/(0.058) =0.517ohm`

Resistance for case b:

`R_(b) =(pL_(b) )/(A_(b) ) =(pL_(0))/(2L_(0)4L_(0) ) =(p)/(8L_(0)) =(1.5x10^(-2) )/(8*0.058) =0.032ohm`

Resistance for case c:

`R_(c) =(pL_(c))/(A_(c) ) =(p2L_(0))/(L_(0)4L_(0)) =(p)/(2L_(0)) =(1.5x10^(-2) )/(2*0.058) =0.129ohm`

b) The current is equal to:

Current for case a:

`I_(a) =(V)/(R_(a) ) =(3)/(0.517) =5.8A`

Current for case b:

`I_(b) =(V)/(R_(b) ) =(3)/(0.032) =93.75A`

Current for case c:

`I_(c) =(V)/(R_(c) ) =(3)/(0.129) =23.2A`