املي بالله
نائبة المدير العام
Physics| Solved problems
physics| solved problems solved problems
Example 1:An elec. lamp 220 volt, 80 Ω resistance takes current of intensity (I). Find the current intensity and the consumed power.
Solution:
I =V/R = 220/80 = 2.75 A
Pw = V.I = 220×2.75 = 605 W
Example 2: An electric D.C. generator of 120 volts feeds an electric furnace of 24 kw power, find the current intensity.
Solution:
I = P_w/V = 24000/120 = 200 A
Example 3: An electric heater works on 220 volts and consumes a current 3A.Find the electric energy consumed through one hour in kwh.
Solution:
E = V .I .T = (220×3)×10-3×1 = 0.66 KWh.
Example 4: what is the resistivity and the electrical conductivity of copper wire of length 25 m, diameter 0.09 cm , and resistance 0.67 ohm.
Solution:
A = πr^2= 3.14× (0.09/(2×100))2 = 6.36×〖10〗^(-7)m2
ρe = RA/L= (o.67×6.36×〖10〗^(-7))/25 = 1.7×〖10〗^(-8)Ω.m
σ = 1/ρ = 1/(1.7×〖17〗^(-8) ) = 5.865×〖10〗^7 Ω-1m-1
physics| solved problems solved problems
Example 1:An elec. lamp 220 volt, 80 Ω resistance takes current of intensity (I). Find the current intensity and the consumed power.
Solution:
I =V/R = 220/80 = 2.75 A
Pw = V.I = 220×2.75 = 605 W
Example 2: An electric D.C. generator of 120 volts feeds an electric furnace of 24 kw power, find the current intensity.
Solution:
I = P_w/V = 24000/120 = 200 A
Example 3: An electric heater works on 220 volts and consumes a current 3A.Find the electric energy consumed through one hour in kwh.
Solution:
E = V .I .T = (220×3)×10-3×1 = 0.66 KWh.
Example 4: what is the resistivity and the electrical conductivity of copper wire of length 25 m, diameter 0.09 cm , and resistance 0.67 ohm.
Solution:
A = πr^2= 3.14× (0.09/(2×100))2 = 6.36×〖10〗^(-7)m2
ρe = RA/L= (o.67×6.36×〖10〗^(-7))/25 = 1.7×〖10〗^(-8)Ω.m
σ = 1/ρ = 1/(1.7×〖17〗^(-8) ) = 5.865×〖10〗^7 Ω-1m-1
التعديل الأخير بواسطة المشرف:
