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Triss [41]
3 years ago
7

what happens to the current in a circuit if the resitance of the components in the circuit is increased​

Physics
1 answer:
Anit [1.1K]3 years ago
7 0

Answer:

The current decreases.

Explanation:

Current and resistance are inversely proportional. The equation connecting current, resistance and voltage is V = IR, where V is voltage, I is current and R is resistance.

Rearranging this equation, you get:

I = \frac{V}{R}

and

R = \frac{V}{I}

If the value of voltage in both equations remains constant, and the value of R decreases, the value of I will increase. Conversely, if in the second equation R = \frac{V}{I} , the value of V remains constant the value of I decreases, then the value of R, resistance will increase.

Thus, it can be seen that the current will decrease as resistance increases and vice versa.

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xenn [34]
The Answer is TRUE. Hope it helps
8 0
3 years ago
What is the period of a simple pendulum 47 cm long (a) on the Earth, and ( b) when it is in a freely falling elevator?
Liula [17]

Answer:

a)1.37 s

b)∞ ( Infinite)

Explanation:

Given that

L= 47 cm              ( 1 m =100 cm)

L= 0.47 m

a)

On the earth :

Acceleration due to gravity = g

We know that time period of the simple pendulum given as

T=2\pi\sqrt{ \dfrac{L}{g_{{eff}}}

Here

g_{eff}= g

Now by putting the values

T=2\pi \times\sqrt{ \dfrac{0.47}{9.81}}

T=1.37 s

b)

Free falling elevator :

When elevator is falling freely then

g_{eff}= 0            ( This is case of weightless motion)

Therefore

T=2\pi\sqrt{ \dfrac{L}{0}

T=∞  (Infinite)

6 0
3 years ago
DESCRIBE THE FORMATION OF THE SOLAR SYSTEM ACCORDING TO THE NEBULAR THEORY
marishachu [46]

When a cloud of gas and dust in space was disturbed, maybe by the explosion of a nearby star.This explosion made waves in space which squeezed the cloud of gas & dust.

8 0
3 years ago
The volume electric charge density of a solid sphere is given by the following equation: The variable r denotes the distance fro
qwelly [4]

Answer:

62.8 μC

Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So,  Q =  ∫∫∫ρdV

Q =  ∫∫∫ρr²sinθdθdrdΦ

Q =  ∫∫∫(0.2r²)r²sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π

So, Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

Q =  ∫∫0.2r⁴[-cosθ]drdΦ

Q =  ∫∫0.2r⁴-[cosπ - cos0]drdΦ

Q =  ∫∫∫0.2r⁴-[-1 - 1]drdΦ

Q =  ∫∫0.2r⁴-[- 2]drdΦ

Q =  ∫∫0.2r⁴(2)drdΦ

Q =  ∫∫0.4r⁴drdΦ

Q =  ∫0.4r⁴dr∫dΦ

Q =  ∫0.4r⁴dr[Φ]

Q =  ∫0.4r⁴dr[2π - 0]

Q =  ∫0.4r⁴dr[2π]

Q =  ∫0.8πr⁴dr

Q =  0.8π∫r⁴dr

Q =  0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

Q = 0.8π[0.025 m⁵]

Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

3 0
2 years ago
Is there a link between number of bulbs and current drawn from the power pack?
Maurinko [17]

Answer:

Yes, there a link between number of bulbs and current drawn from the power pack.

Explanation:

In an Electrical circuit, we have resistors present in that circuit. These resistors can be connected in two ways.

a) Series connection

b) Parallel connection

There is a link or a relationship between number of bulbs and the current drawn from the power pack. This is because the number of bulbs is equivalent to or equal to the number of resistors.

Hence,

a) In a series connection, the link or relationship between the number of bulbs(resistors) is as the number of light bulbs increases, the current in the power pack (circuit) decreases.

b) In a parallel connection, the link or relationship between the number of bulbs(resistors) is as the number of light bulbs increases, the current in the power pack (circuit) increases.

5 0
3 years ago
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