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3 years ago

There is said to be a short circuit when

1 answer:

6 0

A short circuit is said to have occurred when two nodes of a circuit at different voltages are abnormally connected

There is said to be a short circuit when; One part of a parallel circuit has a very low resistance and almost all the current will flow through this part and very little current will flow through the other part

The correct option is option B.

Reason:

A circuit has to be closed or the the current path has to be closed for

current to flow. A short circuit occurs when current in a circuit flows

through an abnormal path, such as a bridge in a connection to a load that

creates a parallel path for current flow.

Almost all the current flows through the parallel path due to the low

resistance in the connection (almost zero) and the proximity to the power

source, thereby preventing current from reaching its intended destination

Therefore, the correct option is option B.;

Learn more here:

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A car accelerates uniformly from rest to a speed of 76.8 mi/h in 7.62 s. Find the constant acceleration of the car. Answer in un

Vedmedyk [2.9K]

Acceleration = (change in speed) / (time for the change)

   =    (76.8 mi/hr)    / (7.62 sec)

   =    10.08 mile/hour-second

That quantity is a perfectly good acceleration, it has the right
dimensions etc, but the question asks for it in units of meter/s²,
so I have to convert it to the equivalent in meter/s².

   (10.8 mile / hr-sec) x (1,609 meter/mile) x (1 hr / 3600 sec)

= (10.8 · 1,609 · 1) / (3,600) meter / second²

=    4.83 meter / second²   if you round all the numbers as you go
or
4.51 meter/sec²    if you use a calculator and don't round anything

6 0

3 years ago

A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/s. The wheel can be considered

mart [117]

Answer:

$\omega_f=12.2954 \,rad.s^{-1}$ i.e.

$Frequency=1.9569\,rev.s^{-1}$

Explanation:

Given:

angular speed, $\omega_i =2\times 2\pi \,rad.s^{-1}$

mass of the disk, $M=4.9\,kg$

radius of the disk, $R=0.2\,m$

mass of the chunk, $m=2.7\,kg$

radius of the chunk, $r=0.04\,m$

We know that the angular momentum is given by:

$L=I.\omega$

and moment of inertia for a disc:

$I=\frac{1}{2} m.r^2$

According to the conservation of angular momentum, the final angular momentum is equal to the initial angular momentum.

$\frac{1}{2} \times M.R^2\times \omega= \frac{1}{2} \times (M.R^2+m.r^2) \omega_f$

$4.9\times 0.2^2\times (2\times 2\pi)=(4.9\times 0.2^2+2.7\times 0.04^2 )\tiems \omega_f$

$\omega_f=12.2954 \,rad.s^{-1}$

$Frequency=1.9569\,rev.s^{-1}$

4 0

4 years ago

A circuit contains a 12 volt battery and two 3 ohm bulbs in series. Calculate the current

OLEGan [10]

If there are two bulbs in series, and each bulb has

resistance = 3 ohms,

then their total resistance in series is 6 ohms.

When the series pair of bulbs is connected across the terminals
of the battery, the current through the circuit is

Current = (voltage) / (resistance)

= (12 volts) / (6 ohms) = 2 amperes

5 0

4 years ago

Suppose Gabor, a scuba diver, is at a depth of 15m. Assume that: The air pressure in his air tract is the same as the net water

s2008m [1.1K]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at

the surface is $\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = 2.5$

The number of moles of gas that must be released is $n= 0.3538\ mols$

Explanation:

We are told from the question that the pressure at the surface is 1 atm and for each depth of 10m below the surface the pressure increase by 1 atm

This means that the pressure at the depth of the surface would be

$P_d = [\frac{15m}{10m} ] (1 atm) + 1 atm$

$= 2.5 atm$

The ideal gas equation is mathematically represented as

$PV = nRT$

Where P is pressure at the surface

V is the volume

R is the gas constant = 8.314 J/mol. K

making n the subject we have

$n = \frac{PV}{RT}$

Considering at the surface of the water the number of moles at the surface would be

$n_s = \frac{P_sV}{RT}$

Substituting $1 atm = 101325 N/m^2$ for $P_s$ , $6L = 6*10^{-3}m^3$ for volume , 8.314 J/mol. K for R , (37° +273) K for T into the equation

$n_s = \frac{(1atm)(6*10^{-3} m^3)}{(8.314J/mol \cdot K)(37 +273)K}$

$= 0.2359 mol$

To obtain the number of moles at the depth of the water we use

$n_d = \frac{P_d V}{RT}$

Where $P_d \ and \ n_d \$ are pressure and no of moles at the depth of the water

Substituting values we have

$n_d = \frac{(2.5)(101325 N/m^2)(6*10^{-3}m^3)}{(8.314 J/mol \cdot K)(37 + 273)K}$

$= 0.5897 mol$

Now to obtain the number of moles released we have

$n = n_d - n_s$

$= 0.5897mol - 0.2359mol$

$=0.3538 \ mol$

The molar concentration at the surface of water is

$[\frac{n}{V} ]_{surface} = \frac{0.2359mol}{6*10^-3m^3}$

$=39.31mol/m^3$

The molar concentration at the depth of water is

$[\frac{n}{V} ]_{15m} = \frac{0.5897}{6*10^{-3}}$

$= 98.28 mol/m^3$

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at the surface is

$\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = \frac{98.28}{39.31} =2.5$

6 0

3 years ago

A clothes dryer uses about 30 amps of current from a 240 volt line. How much power does it use?

Aneli [31]

Not sure if this answer is write or wrong but i think its 4500 watts

8 0

3 years ago

Read 2 more answers