A battery operated DVD player uses 12 DV volts from AA batteries and draws a current of 3 A. Calculate the resistance.

If a voltmeter has a less than ideal resistance, say 1 MΩ, and is used to measure the voltage across a resistor of a comparable

Naddik [55]

Answer:

As the difference between the resistance of voltmeter and the resistance being measured gets reduced the error in the reading of the voltmeter gets increased.

Explanation:

An ideal voltmeter has infinite parallel resistance and because of this it doesn't draw any current from the circuit of measurement which means it will measure the exact voltage across the elements.

But practically speaking, a real voltmeter doesn't has infinite resistance therefore, all the practical voltmeters face loading effect to some extent.

As the difference between the resistance of voltmeter and the resistance being measured gets reduced the error in the reading of the voltmeter gets increased. This is why we want to have a greater value of voltmeter resistance, ideally infinite so that the corresponding error is minimized.

Lets consider the given scenario,

A voltmeter has 1 MΩ parallel resistance and the resistance of of measuring element is 500 kΩ or 0.5 MΩ

lets suppose the supplied voltage is 1 V.

First lets assume that the voltmeter is ideal and it has infinite resistance, so in this case voltmeter will measure a voltage of 1 V across the 0.5 MΩ resistor.

Now consider the loading effect, when we connect the voltmeter across the 0.5 MΩ resistor they both become parallel so the resistance is

R = (1*0.5)/(1+0.5)

R = 0.33 MΩ

As you can see the voltmeter will see a reduced resistance and the corresponding voltage also reduces because resistance and voltage are directly proportional.

Therefore, it is preferred to have a very high parallel resistance of the voltmeter.

8 0

3 years ago

Type the correct answer in each box. Use numerals instead of words.

ser-zykov [4K]

Answer:

6.23 newtons per second?

Explanation:

4 0

3 years ago

You know your mass is 62 kg but when you stand on a bathroom scale in an elevator it says your mass is 77 kg what is the acceler

lbvjy [14]

In a stationary situation, the weight of person is
$W=mg=(62 kg)(9.81 m/s^2) = 608.2 N$
This is the weight "felt" by the scale, which is basically the normal reaction applied by the scale on the person, and which uses the value of g (9.81) as reference to convert the weight (602.8 N) into a mass (62 kg).

When the person is in the elevator, the scale says 77 kg. The scale is still using the same value of conversion (9.81), so the apparent weight "felt" by the scale is
$W' = m'g=(77 kg)(9.81 m/s^2)=755.4 N$
This is the normal reaction applied by the scale on the person, and which is directed upward. Besides this force, there is still the weight W of the person, acting downward. So, if we use Newton's second law:
$\sum F = ma$
$W-W'=ma$
where a is the acceleration of the elevator. If we solve for a, we find
$a= \frac{W-W'}{m}= \frac{608.2N-755.4N}{62 kg}=-2.37 m/s^2$
The negative sign means the acceleration is in the opposite direction of g (which we take positive), so it means the elevator is going upward.

4 0

3 years ago

An electric travel blanket uses a car's 12V supply. The current is 3A.

snow_lady [41]

For electrical devices . . .

   Power dissipated = (voltage) x (current) =

   (12 V) x (3.0 A) = 36 watts .

1 watt means 1 joule per second

   (36 joule/sec) x (60 sec/min) x (10 min) = 21,600 joules

5 0

3 years ago

Two objects with masses m and 3m are connected by a compressed spring so that the energy stored in the spring is 375 joules. If

cluponka [151]

Answer:

281.25 J

Explanation:

We are told that the two objects with masses m and 3m.

Also that energy stored in the spring is 375 joules.

Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3

Thus, from principle of conservation of energy, we have;

½mv² + ½(3m)(v/3)² = 375J

(m + 3m/9)½v² = 375

(4/3)m × ½v² = 375

Multiply both sides by ¾ to get;

½mv² = 375 × ¾

½mv² = 281.25 J

Therefore, energy of lighter body is 281.25 J

7 0

2 years ago