How to find the voltage(B Aab) in series parallel circuit?
1 answer:
Answer:
Vab ≈ 3.426 V
Explanation:
First of all, it is convenient to find the equivalent parallel resistance of R5 and R6. That will be ...
R56 = (R5)(R6)/(R5 +R6) = (1000)(1500)/(1000 +1500) = 600
Then we can call V1 the voltage at the top of R2. The voltage at Va is a divider from V1:
Va = V1·(R4/(R3+R4)) = V1(560/1030) ≈ 0.543689V1
The voltage at Vb is also a divider from V1:
Vb = V1·(R7+R8)/(R2 +R56 +R7 +R8) = V1(780/1710) ≈ 0.456140V1
The parallel branches containing Va and Vb have an effective resistance of ...
(1030)(1710)/(1030+1710) = 642.81
That forms a divider with R1 to give V1:
V1 = (100 V)642.81/(1000 +642.81) ≈ 39.1287 V
The difference Va-Vb is ...
Vab = (39.1287 V)(0.543689 -0.456140) ≈ 3.426 V
_____
We have done this using parallel resistance and voltage divider calculations. You can also do it using node voltage equations. Using the same definition for V1 as above, we have ...
(Vs -V1)/R1 +(Vb -V1)/(R56+R2) +(Va-V1)/R3 = 0
(V1 -Vb)/(R56 +R2) -Vb/(R7+R8) = 0
(V1 -Va)/R3 -Va/R4 = 0
The solution of interest is the value of Vab, shown in the attachment. It computes as 154200/45013 V ≈ 3.42568 V.
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Answer:
2)
a) to the right of the dipole E_total = kq [1 / (r + a)² - 1 / r²]
b)To the left of the dipole E_total = - k q [1 / r² - 1 / (r + a)²]
c) at a point between the dipole, that is -a <x <a
E_total = kq [1 / x² + 1 / (2a-x)²]
d) on the vertical line at the midpoint of the dipole (x = 0)
E_toal = 2 kq 1 / (a + y)² cos θ
Explanation:
2) they ask us for the electric field in different positions between the dipole and a point of interest. Using the principle of superposition.
This principle states that we can analyze the field created by each charge separately and add its value and this will be the field at that point
Let's analyze each point separately.
The test charge is a positive charge and in the reference frame it is at the midpoint between the two charges.
a) to the right of the dipole
The electric charge creates an outgoing field, to the right, but as it is further away the field is of less intensity
E₊ = k q / (r + a)²
where 2a is the distance between the charges of the dipole and the field is to the right
the negative charge creates an incoming field of magnitude
E₋ = -k q / r²
The field is to the left
therefore the total field is the sum of these two fields
E_total = E₊ + E₋
E_total = kq [1 / (r + a)² - 1 / r²]
we can see that the field to the right of the dipole is incoming and of magnitude more similar to the field of the negative charge as the distance increases.
b) To the left of the dipole
The result is similar to the previous one by the opposite sign, since the closest charge is the positive one
E₊ is to the left and E₋ is to the right
E_total = - k q [1 / r² - 1 / (r + a)²]
We see that this field is also directed to the left
c) at a point between the dipole, that is -a <x <a
In this case the E₊ field points to the right and the E₋ field points to the right
E₊ = k q 1 / x²
E₋ = k q 1 / (2a-x)²
E_total = kq [1 / x² + 1 / (2a-x)²]
in this case the field points to the right
d) on the vertical line at the midpoint of the dipole (x = 0)
In this case the E₊ field points in the direction of the positive charge and the test charge
in E₋ field the ni is between the test charge and the negative charge,
the resultant of a horizontal field in zirconium on the x axis (where the negative charge is)
E₊ = kq 1 / (a + y) 2
E₋ = kp 1 / (a + y) 2
E_total = E₊ₓ + E_{-x}
E_toal = 2 kq 1 / (a + y)² cos θ
e) same as the previous part, but on the negative side
E_toal = 2 kq 1 / (a + y)² cos θ
When analyzing the previous answer there is no point where the field is zero
The different configurations are outlined in the attached
3) We are asked to repeat part 2 changing the negative charge for a positive one, so in this case the two charges are positive
a) to the right
in this case the two field goes to the right
E_total = kq [1 / (r + a)² + 1 / r²]
b) to the left
E_total = - kq [1 / (r + a)² + 1 / r²]
c) between the two charges
E₊ goes to the right
E₋ goes to the left
E_total = kq [1 / x² - 1 / (2a-x)²]
d) between vertical line at x = 0
E₊ salient between test charge and positive charge
E_total = 2 kq 1 / (a + y)² sin θ
In this configuration at the point between the two charges the field is zero
Answer:
Explanation:
Given data:
P_1 power = 20 dBm = 0.1 watt
coupling factor is 20dB
Directivity = 35 dB
We know that
coupling factor
solving for final power
Directivity
output Power