Answer:
1.144 A
Explanation:
given that;
the length of the wire = 2.0 mm
the diameter of the wire = 1.0 mm
the variable resistivity R =
Voltage of the battery = 17.0 v
Now; the resistivity of the variable (dR) can be expressed as =
Taking the integral of both sides;we have:
R = 14.863 Ω
Since V = IR
I = 1.144 A
∴ the current if this wire if it is connected to the terminals of a 17.0V battery = 1.144 A
Answer:
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Explanation:
Answer:
The fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is
Explanation:
From the question we are told that
The time constant
The potential across the capacitor can be mathematically represented as
Where is the voltage of the capacitor when it is fully charged
So at
Generally energy stored in a capacitor is mathematically represented as
In this equation the energy stored is directly proportional to the the square of the potential across the capacitor
Now since capacitance is constant at
The energy stored can be evaluated at as
Hence the fraction of the energy stored in an initially uncharged capacitor is