Need help identifying the rest of the elements!

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

#learnwithBrainly

7 0

3 years ago

Mass, size and color are examples of ___

BigorU [14]

The short answer (and the long one for that matter) is physical properties of chemicals. If you are being marked by a machine, likely the answer is going to be physical properties.

6 0

3 years ago

Antonina throws a coin straight up from a height of

vichka [17]

Answer:

$s=vt-\frac{1}{2}gt^2$

Explanation:

We could use the following suvat equation:

$s=vt-\frac{1}{2}gt^2$

where

s is the vertical displacement of the coin

v is its final velocity, when it hits the water

t is the time

g is the acceleration of gravity

Taking upward as positive direction, in this problem we have:

s = -1.2 m

$g=-9.8 m/s^2$

And the coin reaches the water when

t = 1.3 s

Substituting these data, we can find v:

$v=\frac{s}{t}+\frac{1}{2}gt=-\frac{1.2}{1.3}+\frac{1}{2}(-9.8)(1.3)=-7.3 m/s$

where the negative sign means the direction is downward.

5 0

3 years ago

A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th

Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(4)}{2}(2(0)+4)$

$F = 25000 lb$

b) Force on deep end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(10)}{2} (2(0)+10)$

$F = 187500 lb$

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

$F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)$