Answer:
t = 5.48 × 10⁻³ s
Explanation:
Given:
ΔV = ΔVmax × sin(2πft)
frequency, f = 16.9Hz
thus,
ΔV = ΔVmax × sin(2π×16.9×ft)
Now,
Let 'R' be the resistance
Also according to the ohms law
i = V/R
where,
i = current
V = voltage
hence,

also, given at time 't' the current in the circuit is 55.0% of the peak current
thus

thus,
or
or
or
t = 5.48 × 10⁻³ s (Answer)
Answer:
Option C
Explanation:
According to the formula
So
If we use wide wire we increase the area of cross section so resistance decreases
<h2>
Answer:</h2>
(a) 3.96 x 10⁵C
(b) 4.752 x 10⁶ J
<h2>
Explanation:</h2>
(a) The given charge (Q) is 110 A·h (ampere hour)
Converting this to A·s (ampere second) gives the number of coulombs the charge represents. This is done as follows;
=> Q = 110A·h
=> Q = 110 x 1A x 1h [1 hour = 3600 seconds]
=> Q = 110 x A x 3600s
=> Q = 396000A·s
=> Q = 3.96 x 10⁵A·s = 3.96 x 10⁵C
Therefore, the number of coulombs of charge is 3.96 x 10⁵C
(b) The energy (E) involved in the process is given by;
E = Q x V -----------------(i)
Where;
Q = magnitude of the charge = 3.96 x 10⁵C
V = electric potential = 12V
Substitute these values into equation (i) as follows;
E = 3.96 x 10⁵ x 12
E = 47.52 x 10⁵ J
E = 4.752 x 10⁶ J
Therefore, the amount of energy involved is 4.752 x 10⁶ J
Answer:
a)η = 69.18 %
b)W= 1210 J
c)P=3967.21 W
Explanation:
Given that
Q₁ = 1749 J
Q₂ = 539 J
From first law of thermodynamics
Q₁ = Q₂ +W
W=Work out put
Q₂=Heat rejected to the cold reservoir
Q₁ =heat absorb by hot reservoir
W= Q₁- Q₂
W= 1210 J
The efficiency given as



η = 69.18 %
We know that rate of work done is known as power


P=3967.21 W