In the following circuit (Fig.3), calculate the intensity I through the resistance 3 using the principle of superposition.

For the circuit shown, calculate

RSB [31]

For the circuit shown

a.the total resistance = R = 35.9 Ω

b. when total current = 2 A, then total voltage = V = 71.8 V

c.the current through resistor of resistance 56Ω = I₁ = 1.28 A

the current through resistor of resistance 100Ω = I₂ = 0.72 A

Explanation:

a)

Resistors can be connected in series or in parallel. For series combination Resultant resistance of the circuit is given by

R = R₁ + R₂ + R₃ +...........+Rₙ

But is our case as shown in the picture, Both the resistors are connected in parallel and for parallel combination resultant, resultant resistance of the circuit is given by

(1 / R) = (1 / R₁) + (1 / R₂) + (1 / R₃) +.......+ (1 / Rₙ)

So

1 / R = (1 / R₁) + (1 / R₂)

Let

R₁ = 56Ω and R₂ = 100Ω

1 / R = 1/56 + 1/100

1 / R = 39 / 1400

R = 1400 / 39

R = 35.9 Ω

b)

Part b can be solved by Ohm's Law Which is stated as "The current flowing through a circuit is directly proportional to the potential difference across its ends provided the physical state such as temperature of the conductor (circuit) does not change."

Mathematically

V ∝ I

V = IR

Where V is the potential difference across the ends of the conductor and I is the Current flowing through the circuit. R is the resistance of the circuit.

Given data:

Total current = I = 2 A

Resistance = R = 35.9 Ω

Voltage = ?

By Ohm's Law

V = IR

V = 2*35.9

V = 71.8 V

c)

To solve current through each resistor, We can use current division formula which is given as

I₁ = I[ R₂ / (R₁ + R₂) ]

I₂ = I[ R₁ / (R₁ + R₂) ]

Also we can take Voltage across each resistor equal to V = 71.8 V because Potential difference across each resistors connected in parallel remain same . And by using Ohm's law divide the value of potential difference with the value of respective resistor to find the current through each resistors.

By Ohm's Law

V = IR

I = V / R

I₁ = 71.8 / 56

I₁ = 1.28 A

I₂ = V / R

I₂ = 71.8 / 100

I₂ = 0.72 A

Learn more about Resistors and Ohm's law from

https://brainly.in/question/9744300

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7 0

3 years ago

The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is this case is 0 as the road is level

Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0

3 years ago

A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi

Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

$d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L$

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

$x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L$