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

Answer:

a) Ra = 0.517 Ω

Rb = 0.032 Ω

Rc = 0.129 Ω

b) Ia = 5.8A

Ib = 93.75A

Ic = 23.2 A

Explanation:

a) The resistance is equal to:

Resistance for case a:

$R_{a} =\frac{pL_{a} }{A_{a} } =\frac{p*4*L_{0} }{2L_{0}*L_{0} } =\frac{2p}{L_{0} }$

Where

p = 1.5x10⁻²Ωm

L0 = 5.8 cm = 0.058 m

$R_{a} =\frac{2*1.5x10^{-2} }{0.058} =0.517ohm$

Resistance for case b:

$R_{b} =\frac{pL_{b} }{A_{b} } =\frac{pL_{0}}{2L_{0}4L_{0} } =\frac{p}{8L_{0}} =\frac{1.5x10^{-2} }{8*0.058} =0.032ohm$

Resistance for case c:

$R_{c} =\frac{pL_{c}}{A_{c} } =\frac{p2L_{0}}{L_{0}4L_{0}} =\frac{p}{2L_{0}} =\frac{1.5x10^{-2} }{2*0.058} =0.129ohm$

b) The current is equal to:

Current for case a:

$I_{a} =\frac{V}{R_{a} } =\frac{3}{0.517} =5.8A$

Current for case b:

$I_{b} =\frac{V}{R_{b} } =\frac{3}{0.032} =93.75A$

Current for case c:

$I_{c} =\frac{V}{R_{c} } =\frac{3}{0.129} =23.2A$