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Dmitry_Shevchenko [17]
3 years ago
9

Two lightbulbs have powers P1 and P2 when separately connected across an ideal battery with potential difference ΔV. The bulbs a

re then connected in series across the same battery. Determine the total power consumption of the bulbs in this configuration. (Use any variable or symbol stated above as necessary.)
Physics
1 answer:
jarptica [38.1K]3 years ago
6 0

Answer:

P'=\dfrac{P_1P_2}{P_1+P_2}

Explanation:

Lets take

Resistance of bulb 1 =R₁

Resistance of bulb 2 =R₂

As we know that power P

P= ΔV²/R

Given that voltage difference is same for both bulbs

So

P₁R₁= ΔV²       --------1

P₂R₂= ΔV²   -----------2

When these resistance are connected in series then equivalent resistance R

R=R₁+ R₂

The new power P'

P'=ΔV²/R

P'R=ΔV²   ------3

From equation 1 ,2 and 3

P'(R₁+ R₂) = ΔV²

P'\left(\dfrac{\Delta V^2}{P_1}+\dfrac{\Delta V^2}{P_2}\right)=\Delta V^2

P'=\dfrac{P_1P_2}{P_1+P_2}

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What is the voltage across each resistor?
kogti [31]

Answer:

see solution below

Explanation:

The given resistors are connected in series.

Equivalent resistance in series = 30 + 55 + 15

Equivalent resistance in series Rt = 100 ohms

Since the potential difference in the circuit = 36V

Get the current in the circuit first

I = V/Rt

I = 36/100

I = 0.36A

Get the voltage across 30ohms resistor;

V30 = 0.36 * 30

V30  = 10.8volts

Hence the voltage across the 30ohms resistor is 10.8volts

Get the voltage across 55ohms resistor;

V55 = 0.36 * 55

V55  = 19.8volts

Hence the voltage across the 55ohms resistor is 19.8volts

Get the voltage across 15ohms resistor;

V15 = 0.36 * 15

V15  = 5.4volts

Hence the voltage across the 15ohms resistor is 5.4volts

4 0
2 years ago
Amount of force exerted on an object due to gravity is called
MAXImum [283]
The gravitational force between a mass and the Earth is the object'sweight. Mass is considered a measure of an object's inertia, and its weight is the force exerted on the object in a gravitational field. On the surface of the Earth, the two forces are related by the acceleration due to gravity: Fg = mg.

Hoped this helped!
6 0
3 years ago
A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are
Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, \Sigma F_y = 0

The vertical component of the tension, T_y, in each half of the rope is given as follows;

T_y = T × sin(θ)

∴ \Sigma F_y = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

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tino4ka555 [31]

Answer:

he will use kinetic energy to break through the door

Explanation:

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iVinArrow [24]

Answer:

you start sweating and it would get hot

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