All categories

3 years ago

You measure the voltage difference of a circuit to be 5 V and the resistance to be 1,000 ohms.

Physics

1 answer:

Alja [10]3 years ago

8 0

Answer:

The current in the circuit will be $5\times 10^{-3}\ Ampere$ .

Explanation:

Given:

Voltage difference = V = 5 V

Resistance = R = 1000 ohms

To Find:

Current, I = ?

Solution:

Ohm's Law:

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

i.e I ∝ V

∴ V = I × R

Where, I = current

V = voltage

R = resistance

Substituting the values of V and R we get

∴ 5 = I × 1000

$I = \frac{5}{1000} \\I=0.005\ Ampere\\I=5\times 10^{-3}\ Ampere$

Therefore, the current in the circuit will be $5\times 10^{-3}\ Ampere$ .

You might be interested in

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is: