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2 years ago

How much current is in a circult that includes a 9.0-volt battery and a bulb with a resistance of 4.0 ohms?

Physics

1 answer:

SVETLANKA909090 [29]2 years ago

6 0

Answer:

Current, I = 2.3 A

Explanation:

We have,

Voltage of the battery in a circuit is 9 volts

Resistance of the circuit is 4 ohms

It is required to find the current in a circuit. When the voltage and the resistance of the circuit is given then we can find the current in it is given by Ohm's law as :

$V=IR$

I is electric current

$I=\dfrac{V}{R}\\\\I=\dfrac{9}{4}\\\\I=2.25\ A$

I = 2.3 A

So, the current in the circuit is 2.3 A.

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In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phas

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Answer:

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Explanation:

From the question we are told that

The distance between the slits is $d = 0.2 \ mm = \frac{0.2}{1000} = 0.2 *10^{-3} \ m$

The distance to the screen is $D = 100 cm = \frac{100}{100} = 1 \ m$

The wavelength is $\lambda = 400nm$

The distance of the wave from the central maximum is $L = 5mm = 5*10^{-3} m$

Generally the path difference of this waves is mathematically represented as

$y = d sin \theta$

Here $\theta$ is the angle between the the line connecting the mid-point of the slits with the screen and the line connecting the mid-point of the slits to the central maximum

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=> $\theta = tan ^{-1} \frac{L}{D}$

$\theta = tan ^{-1} [\frac{5*10^{-3}}{1}]$

$\theta =0.2865$

Substituting values into the formula for path difference

$y = 0.2 *10^{-3} sin(0.2864)$

$y = 9.997*10^{-7} \ m$

The phase difference is mathematically represented as

$\Delta \phi = \frac{2 \pi }{\lambda } * y$

Substituting values

$\Delta \phi = \frac{2 \pi }{400 *10^{-9} } \ * 9.997*10^{-7}$

$\Delta \phi =5 \pi$

Converting to degree

$\Delta \phi =5 \pi radians = 5 (180^o) = 180^o$

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