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1. Koju bi brzinu postigla kuglica mase 5g i naboja 5μC, ako bi se gibala s mjesta potencijala
30000V na 3000V?


m = 5 g = 5 ⋅ 10 −3 kg
Q = 5µC = 5 ⋅ 10 −6 C
ϕ1 = 30000V
ϕ 2 = 3000V
v=?
     m ⋅ v2
Ek =
       2
U = ∆ϕ
U = 27000V
W = Ek
W = U ⋅Q
        m ⋅ v2
U ⋅Q =
          2
     2 ⋅U ⋅ Q
v2 =
        m
     2 ⋅U ⋅ Q
v=
         m
v = 3 6ms −1
v ≈ 7.35ms −1
2. Kapacitet kondenzatora je 1.3 pF. Ako udvostručimo razmak među pločama i između njih
stavimo vosak kapacitet će biti 2.6 pF. Kolika je relativna permitivnost voska?


C1 = 1.3 pF
C2 = 2.6 pF
εr = ?
C1 1
    =
C2 2
2C1 = C2
2d1 = d 2
                  S
C = ε0 ⋅ εr ⋅
                  d
           S               S
2 ⋅ ε0 ⋅      = ε0 ⋅ εr ⋅
           d1             2d1
            1
2 = εr ⋅
            2
εr = 4
3. Paralelni spoj od 54 žarulja otpora 800Ω priključen je na napon od 220V preko vodiča
otpora 1 Ω. Koliki je napon na svakoj žarulji?

N = 54
Ržarulje = R1 = ... = R54 = 800Ω
Rvodičo = 1Ω
           800Ω
Ržarulja =      ≈ 14.82Ω
            54
Ruk = 14.82Ω + 1Ω = 15.82Ω
U uk = 220V
U uk = I uk ⋅ Ruk
                       220V
I uk = I žarulja =           = 13.9 A
                      15.82Ω
U žarulje = U žarulja
U žarulja = Ržarulja ⋅ I žarulja

U žarulje ≈ 206V
4. Kolika je snaga električnog uređaja priključenog na 220V, ako se 1L vode ugrije od 20°C
do temperature vrenja za 3.5min.

U = 220V
V = 1L
∆T = 80 K
t = 3.5 min = 210 s
Q =W = ?
ρ = 1kgL−1
m = 1kg
c = 4190 J ( kgK )
                           −1


     Q
c=
   m∆T
Q = 335200 J
P=?
   W
P=   ≈ 1596.19W
   t
5. 3 otpornika otpora 20Ω, 40Ω i 60Ω spojena su paralelno. Jakost struje kroz otpornik s
najmanjim otporom je 6A. Izračunaj jakost struje kroz ostale otpornike i ukupnu jakost u
krugu.

R1 = 20Ω
R2 = 40Ω
R3 = 60Ω
I1 = 6 A
I 2,3,uk = ?
     U uk
I=
      R
U uk = stalan = 120V
I 2 = 3A
I3 = 2A
I uk = I 1 + I 2 + 3 = 11A

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Ispit iz fizike

  • 1. 1. Koju bi brzinu postigla kuglica mase 5g i naboja 5μC, ako bi se gibala s mjesta potencijala 30000V na 3000V? m = 5 g = 5 ⋅ 10 −3 kg Q = 5µC = 5 ⋅ 10 −6 C ϕ1 = 30000V ϕ 2 = 3000V v=? m ⋅ v2 Ek = 2 U = ∆ϕ U = 27000V W = Ek W = U ⋅Q m ⋅ v2 U ⋅Q = 2 2 ⋅U ⋅ Q v2 = m 2 ⋅U ⋅ Q v= m v = 3 6ms −1 v ≈ 7.35ms −1
  • 2. 2. Kapacitet kondenzatora je 1.3 pF. Ako udvostručimo razmak među pločama i između njih stavimo vosak kapacitet će biti 2.6 pF. Kolika je relativna permitivnost voska? C1 = 1.3 pF C2 = 2.6 pF εr = ? C1 1 = C2 2 2C1 = C2 2d1 = d 2 S C = ε0 ⋅ εr ⋅ d S S 2 ⋅ ε0 ⋅ = ε0 ⋅ εr ⋅ d1 2d1 1 2 = εr ⋅ 2 εr = 4
  • 3. 3. Paralelni spoj od 54 žarulja otpora 800Ω priključen je na napon od 220V preko vodiča otpora 1 Ω. Koliki je napon na svakoj žarulji? N = 54 Ržarulje = R1 = ... = R54 = 800Ω Rvodičo = 1Ω 800Ω Ržarulja = ≈ 14.82Ω 54 Ruk = 14.82Ω + 1Ω = 15.82Ω U uk = 220V U uk = I uk ⋅ Ruk 220V I uk = I žarulja = = 13.9 A 15.82Ω U žarulje = U žarulja U žarulja = Ržarulja ⋅ I žarulja U žarulje ≈ 206V
  • 4. 4. Kolika je snaga električnog uređaja priključenog na 220V, ako se 1L vode ugrije od 20°C do temperature vrenja za 3.5min. U = 220V V = 1L ∆T = 80 K t = 3.5 min = 210 s Q =W = ? ρ = 1kgL−1 m = 1kg c = 4190 J ( kgK ) −1 Q c= m∆T Q = 335200 J P=? W P= ≈ 1596.19W t
  • 5. 5. 3 otpornika otpora 20Ω, 40Ω i 60Ω spojena su paralelno. Jakost struje kroz otpornik s najmanjim otporom je 6A. Izračunaj jakost struje kroz ostale otpornike i ukupnu jakost u krugu. R1 = 20Ω R2 = 40Ω R3 = 60Ω I1 = 6 A I 2,3,uk = ? U uk I= R U uk = stalan = 120V I 2 = 3A I3 = 2A I uk = I 1 + I 2 + 3 = 11A