Answer:
6.88 mA
Explanation:
Given:
Resistance, R = 594 Ω
Capacitance = 1.3 μF
emf, V = 6.53 V
Time, t = 1 time constant
Now,
The initial current, I₀ = 
or
I₀ = 
or
I₀ = 0.0109 A
also,
I = ![I_0[1-e^{-\frac{t}{\tau}}]](https://tex.z-dn.net/?f=I_0%5B1-e%5E%7B-%5Cfrac%7Bt%7D%7B%5Ctau%7D%7D%5D)
here,
τ = time constant
e = 2.717
on substituting the respective values, we get
I = ![0.0109[1-e^{-\frac{\tau}{\tau}}]](https://tex.z-dn.net/?f=0.0109%5B1-e%5E%7B-%5Cfrac%7B%5Ctau%7D%7B%5Ctau%7D%7D%5D)
or
I = ![0.0109[1-2.717^{-1}]](https://tex.z-dn.net/?f=0.0109%5B1-2.717%5E%7B-1%7D%5D)
or
I = 0.00688 A
or
I = 6.88 mA
1. b or c
2. c
3. a? or d
4.
5. a
i think its d im not sure
Answer:
E. two times the original diameter
Explanation:
Resistance of a wire is:
R = ρ L/A
where ρ is the resistivity of the material, L is the length, and A is the cross-sectional area.
For a round wire with diameter d:
R = ρ L / (¼ π d²)
The two wires must have the same resistance, so:
ρ₁ L₁ / (¼ π d₁²) = ρ₂ L₂ / (¼ π d₂²)
The wires are made of the same material, so ρ₁ = ρ₂:
L₁ / (¼ π d₁²) = L₂ / (¼ π d₂²)
The new length is four times the old, so 4 L₁ = L₂:
L₁ / (¼ π d₁²) = 4 L₁ / (¼ π d₂²)
1 / (¼ π d₁²) = 4 / (¼ π d₂²)
Solving:
1 / (d₁²) = 4 / (d₂²)
(d₂²) / (d₁²) = 4
(d₂ / d₁)² = 4
d₂ / d₁ = 2
So the new wire must have a diameter twice as large as the old wire.
