You can do 50 and 10 and carry the 30 and it would be in the same power so you’ll have the same energy
Answer:
a. Zin = 41.25 - j 16.35 Ω
b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶
c. Pin = 216 w
d. PL = Pin = 216 w
e. Pg = 478.4 w , Pzg = 262.4 w
Explanation:
a.
Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]
βl = 2π / λ * 0.15 λ = 54 °
Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]
Zin = 41.25 - j 16.35 Ω
b.
I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶
V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)
V₁ = 143. 6 e⁻ ¹¹ ⁴⁶
c.
Pin = ¹/₂ * Re * [V₁ * I₁]
Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)
Pin = 216 w
d.
The power PL and Pin are the same as the line is lossless input to the line ends up in the load so
PL = Pin
PL = 216 w
e.
Pg Generator
Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)
Pg = 478.4 w
Pzg dissipated
Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50
Pzg = 262.4 w
Answer:
The peak emf of the generator is 40.94 V.
Explanation:
Given that,
Number of turns in primary coil= 11
Number of turns in secondary coil= 18
Peak voltage = 67 V
We nee to calculate the peak emf
Using relation of number of turns and emf
Where, N₁ = Number of turns in primary coil
N₂ = Number of turns in secondary coil
E₂ = emf across secondary coil
Put the value into the formula
Hence, The peak emf of the generator is 40.94 V.
Answer:
The water level in the bath tub is rising at a rate of 0.0111 ft/s
Explanation:
Volume of the bath tub = (Area of base) × (height)
Area of base = 18 ft² (constant)
Height = h (variable)
V = 18h
(dV/dt) = 18 (dh/dt)
If (dV/dt) = 0.2 ft³/s
0.2 = 18 (dh/dt)
(dh/dt) = (0.2/18)
(dh/dt) = 0.0111 ft/s
Hope this Helps!!!
Answer:
1. 150C.
2. 50sec
3.1.5a
Explanation:
1. I = Q/T
Q= 30x5
=150c
2.applying the formulae, I = Q/T
T= Q/I
=500/10
=50sec.
3. using the formulae i=q/t
i= 120/80
=1.5a.
