Question

Este o varianta de bac 2009 electricitate

Answer 1

Se da:
E=4,5V
U=4V
P=2W
a)R1=?ohm
b)r=?ohm
c)p1=p2; S1=S2; 2*l1=l2; P2=?W
d)t=2min=120s; Q3=?J

Formule:
a)P=U²/R1⇒R1=U²/P

b)I=E/(R+r)
U/R=E/(R+r)
R+r=R*E/U
r=(R*E/U)-R
r=R*[(E/U)-1]

c)P2=I2²*R2

R2/R1=(p2l2/S2)/(p1l1/S1)
R2/R1=p2*2l1*S1/p1*l1*S2
R2/R1=2
R2=2*R1

I2=E/(R2+r)
I2=E/(2*R1+r)

P2=2*R1*E²/(2*R1+r)²

d)Nu se prea intelege energia cui, circuitului extern; intern sau generatorului, dar cred ca a celui extern.

Q3=I3²*R3*t

1/R3=(1/R1)+(1/R2)
1/R3=(1/R1)+(1/2*R1)
1/R3=3/2*R1
R3=2*R1/3

I3=E/(R3+r)
I3=E/[(2*R1/3)+r]

Q3=2*R1*t*E²/3*[(2*R1/3)+r]²

Calcule:
a)R1=4²/2=8ohm
b)r=8*[(4,5/4)-1]=1ohm
c)P2=2*8*4,5²/(2*8+1)²≈1,12W
d)Q3=2*8*120*4,5²/3*[(2*8/3)+1]²≈323,1J