Question

a) Calculate the power dissipated in R₃.

R₁ = 329 Ω
R₂ = 879 Ω
R₃ = 622 Ω
R₄ = 118 Ω
V = 6.0 V

F: R

$prob02_twocylresist.gif$

The picture shows a battery connected to two wires in parallel. Both wires are made of the same material and have the same length, but the diameter of wire A is twice the diameter of wire B.
For each statement select T True or F False.

a)The current through the battery is four times the current through wire A.

b)The voltage drop across wire A is larger than the voltage drop across wire B.

c)The resistance of wire A is one quarter of the resistance of wire B.

d)The resistance of wire B is twice as large as the resistance of wire A.

e)The power dissipated in wire A is 16 times the power dissipated in wire B.

Answer 1

The total resistance is

$R=R_3+R_4+\frac{R_1 R_2}{R_1+R_2}=979.397\,\,\Omega$

The current through the battery is

$I=V/R = 6/979.397=0.00613\,\,A$

The power dissipated in R3 is

$P=I^2 R_3 =0.00613^2\times 622=0.0233\,\,W$