An electric current is created in a long thin wire. How will increasing the current and changing the direction of the current ef
2 answers:
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Answer:
Part a)
Part b)
Explanation:
As we know that the energy of capacitor when it is not connected to potential source is given as
As we know that initial energy is given as
now we know that capacitance of parallel plate capacitor is given as
now the new capacitance when distance is changed from 3.80 mm to 1.45 mm
Now the new energy of the capacitor is given as
Part b)
Now if the voltage difference between the plates of capacitor is given constant
now the energy energy of capacitor is
now when capacitance is changed to new value then new energy is given as
Answer:
The final temperature of the element = 262.67°C
The power dissipated in the heating element initially = 163.21 W
The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W
Explanation:
Our given parameters include;
A Nichrome heating element operates on 120 V.
Voltage (V) = 120V
Initial Current (I₁) = 1.36 A
Initial Temperature (T₁) = 28°C
Final Current (I₂) = 1.23 A
Final Temperature (T₂) = unknown ????
Temperature dependencies of resistance is given by:
---------------------- (1)
in which R₁ is the resistance at temperature T₁
is the resistance at temperature T₂
Given that V= IR
R =
Therefore, the resistance at temperature 28°C is;
= 88.24Ω
=
= 97.56Ω
From (1) above;
97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]
1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)
0.1056 = 4.5×10⁻⁴(T₂-28°C)
T - 28° C = 234.67
T = 234.67 + 28° C
T = 262.67 ° C
(b)
What is the power dissipated in the heating element initially and when the current reaches 1.23 A
The power dissipated in the heating element initially can be calculated as:
P = I²₁R₂₈
P = (1.36A)²(88.24Ω)
P = 163.209 W
P ≅ 163.21 W
The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:
P = (1.23)²(97.56Ω)
P = 147.598524
P ≅ 147.60 W