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2 years ago

13

What is the power through a device with a resistance of 100 ohms if a current of 8 A is running through it?

1 answer:

Kazeer [188]2 years ago

5 0

Answer:

6400 W (or) 6.4 KW

Explanation:

Formula we use,

→ P = I²R

Let's solve for the power of device,

→ P = I²R

→ P = (8)² × 100

→ P = 64 × 100

→ [ P = 6400 W ]

Hence, the power is 6400 W.

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A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the pot

denpristay [2]

Answer:

Part a)

V = 18.16 V

Part b)

$P_r = 345 Watt$

Part c)

P = 672 Watt

Part d)

V = 5.84 V

Part e)

$P_r = 345 Watt$

Explanation:

Part a)

When battery is in charging mode

then the potential difference at the terminal of the cell is more than its EMF and it is given as

$\Delta V = E + i r$

here we have

$E = 12 V$

$i = 56 A$

$r = 0.11$

now we have

$\Delta V = 12 + (0.11)(56) = 18.16 V$

Part b)

Rate of energy dissipation inside the battery is the energy across internal resistance

so it is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

Part c)

Rate of energy conversion into EMF is given as

$P_{emf} = i E$

$P_{emf} = (56)(12)$

$P_{emf} = 672 Watt$

Now battery is giving current to other circuit so now it is discharging

now we have

Part d)

$V = E - i r$

$V = 12 - (56)(0.11)$

$V = 12 - 6.16 = 5.84 V$

Part e)

now the rate of energy dissipation is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

7 0

3 years ago

State coulombs law in word

amid [387]

<h2><em>state coulombs law in word</em></h2>

<em>: a statement in physics: <u>the force of attraction or repulsion acting along a straight line between two electric charges is directly proportional to the product of the charges and inversely to the square of the distance between </u></em><em><u>them</u></em>

<em><u>hope </u></em><em><u>it</u></em><em><u> helps</u></em>

<em><u>#</u></em><em><u>c</u></em><em><u>a</u></em><em><u>r</u></em><em><u>r</u></em><em><u>y</u></em><em><u> </u></em><em><u>on</u></em><em><u> learning</u></em>

3 0

3 years ago

Why is it that the two possible faults with inductors are short circuit and open circuit?

lukranit [14]

These are the most common type of faults not just inductors but also with other elements too like resistors,transformers, generators etc.
open circuit fault means the flow of current is disrupted some how in the circuit and the circuit stops operating. and for short circuit fault the current in the system will be pretty high and this short circuit current or fault current will always run back to the fault location, if the inductor got short circuited somehow then the fault current will only run through it because it will then provide a very low impedence path

5 0

3 years ago

There. That is better.

mihalych1998 [28]

a
a
b
b
a
b
a
This will really help you learn a lot.

6 0

4 years ago

With detailed explaniation

belka [17]

Ø=37°
Initial velocity=u=20m/s
g=10m/s²

#A

$\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}$

$\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}$

$\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}$

$\\ \rm\Rrightarrow H_{max}=20sin^237$

$\\ \rm\Rrightarrow H_{max}=7.2m$

#B

$\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}$

$\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}$

$\\ \rm\Rrightarrow R=40sin74$

$\\ \rm\Rrightarrow R=38.5m$

#C

$\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}$

$\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}$

$\\ \rm\Rrightarrow T=4sin37$

$\\ \rm\Rrightarrow T=2.4s$

Now

$\\ \rm\Rrightarrow v=u-gt$

$\\ \rm\Rrightarrow v=20-10(2.4)$

$\\ \rm\Rrightarrow v=20-24$

$\\ \rm\Rrightarrow v=-4m/s$

4 0

2 years ago