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olchik [2.2K]
3 years ago
13

What component is used to measure the current in a circuit

Physics
2 answers:
ELEN [110]3 years ago
7 0

Answer:

connect an anmeter in series with the resistor

Explanation:

apex

iragen [17]3 years ago
3 0
Measuring current. A device called an ammeter is used to measure current. Some types of ammeter have a pointer on a dial, but most have a digital readout. To measure the current flowing through a component in a circuit, you must connect the ammeter in series with it.
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What are possible units for impulse? Check all that apply. kg • m kg • N • s N • m
Vlada [557]

We know that impulse is simply the product of Force and time:

Impulse = Force * time

 

Since Force has a unit of Newton or kg m/s^2 and time is in seconds, therefore impulse can have units as:

N s

or

<span>kg m/s</span>

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3 years ago
How does the force of gravity between two bodies change when the distance between them doubles? 1. unable to determine; the mass
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6. Drop to one quarter of its original value
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3 years ago
An overhead electric power line carries a maximum current of 125 A. What is the magnitude of the maximum magnetic field at a poi
BARSIC [14]

Answer:

B= 55.6×10^(-7) Tesla

Explanation:

B= μoI/(2πr)

B:  magnetic field strength

μo: permeability of free space and is equal to  4π×10^(-7) T.m/A

r: distance from the wire

I : current in the wire

B= (4π×10^(-7)×125)/(2π×4.5)

B= 55.6×10^(-7) Tesla

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4 years ago
Creates an image that appears upside down behind the focal point
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An image that appears upside down behind the focal point is an image that is reflected on a concave mirror. Mirrors reflect different kinds of images based on the placement of an object that is reflected towards it. There are two kinds of mirrors, concave and a convex mirrors, the latter makes objects seem smaller and farther than where it is exactly.
8 0
3 years ago
When a certain air-filled parallel-plate capacitor is connected across a battery, it acquires a charge of magnitude 172 μC on ea
crimeas [40]

Answer:

k = 2.279

Explanation:

Given:

Magnitude of charge on each plate, Q = 172 μC

Now,

the capacitance, C of a capacitor is given as:

C = Q/V

where,

V is the potential difference

Thus, the capacitance due to the charge of 172 μC will be

C = \frac{(172\ \mu C)}{V}

Now, when the when the additional charge is accumulated

the capacitance (C') will be

C' = \frac{(172+220)\ \mu C)}{V}

or

C' = \frac{(392)\ \mu C)}{V}

now the dielectric constant (k) is given as:

k=\frac{C'}{C}

substituting the values, we get

k=\frac{\frac{(392\ \mu C)}{V}}{\frac{(172)\ \mu C)}{V}}

or

k = 2.279

6 0
4 years ago
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