Question

Consider the following (a) A rectangular block has the dimensions lxwxh. When a battery with a voltage is connected across the end faces as shown in the left figure above, a current Inows through the circut If the same battery is connected across the top and bottom faces of the block as shown in the right figure, what is the current I, that flows through the circuit? Give your answer in terms of (Use the following as necessary: t, w, , - (4) X What is the resistance for the two case? How does Ohm's law relate current to voltage (b) suppose I 400 cm, w * 2.2 cm, and h - 2.00 cm, the battery voltage is 1.5 Vand I, 1.10 A determine the value of lo

Answer 1

so ما + (ca case-1.) According to Ohm's law Where, 으 I, RI --- Va voltage of the batt attery I a enrrent in the cirenit Ri= resistance of the cirunit Now is l Where A block =) Ri=el resistance of this circuit R giren by R, 2 t. f f = resistivity of the la length of the Condrefor A= cross-sectional area of the conductor I wxh Wh Putting the value of R in egination ☺ we get vi I, X el Wh

Lady two (Case-2) for this and case where V = [2 R2 R₂ is h R₂2 Pi I a enrre rent in the ciremit R₂ = resistance of the conductor given by ve voltage of the battery (same (Wxl) ha length of the condneter for and case (wx l) a cross-section area of the conducts fa resistivity of the conductor (same ) Putting the value of R2 in equation th (we) comparing equation we + ge a va 12 x ۱۷ Co and equation we w get, I, X el f₂ X 8h Wh (ul) 12 I, hr 2 12 le XI ch

e = 4.00 em x I h 2.00 cm 12 = () * => 1 = (4.323.104 Il - 3,10 ☺ A :-) I 2 2 (4 x 3,10 ) 12 12,4 A 12.4 value of I2 = .