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3 years ago

To measure the current through a resistor, a _ must be placed in _ with the resistor.

Engineering

1 answer:

fgiga [73]3 years ago

5 0

Answer:

Explanation:

To measure the current through a resistor, the Ammeter must be placed in series with the resistor so that the same current that passes through the resistor also passes through the Ammeter. Ammeter is a measuring device used to measure current.

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An isentropic steam turbine processes 2 kg/s of steam at 3 MPa, which is exhausted at50 kPa and 100C. Five percent of this flow

borishaifa [10]

Answer:

2285kw

Explanation:

since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.

Mathematically,

$\\ E_{inflow} = E_{outflow}$

Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy $\\ E_{inflow} = m_{1} h_{1}$ .

$\\$

$E_{outflow} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$

$\\$

Where $m_{1} h_{1}$ are the mass flow rate and the enthalpies at the inlet at a pressure of 3Mpa $\\$ ,

$m_{2} h_{2}$ are the mass flow rate and the enthalpies at the outlet 2 where we have a pressure of 500kpa respectively. $\\$ ,

and $m_{3} h_{3}$ are the mass flow rate and the enthalpies at the outlet 3 where we have a pressure of 50kpa respectively. $\\$ ,

We can now express write out the required equation by substituting the new expression for the energies $\\$

$m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$

from the above equation, the unknown are the enthalpy values and the mass flow rate. $\\$

first let us determine the enthalpy values at the inlet and the out let using the Superheated water table. $\\$

It is more convenient to start from outlet 3 were we have a temperature $100^{0}C$ and pressure value of (50kpa or 0.05Mpa ). using double interpolation method on the superheated water table to determine the enthalpy value with careful calculation we have $\\$

$h_{3}$ = 2682.4 KJ/KG , at this point also from the table the entropy value , $s_{3}$ value is 7.6953 KJ/Kg.K. $\\$

Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy is constant at all inlet and outlet. $\\$

So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of 7.6953 KJ/Kg.K with careful interpolation we arrive at a enthalpy value of 3206.5KJ/Kg. $\\$

Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K we arrive at enthalpy value of 3851.2KJ/Kg. $\\$

Now we determine the mass flow rate at each inlet and outlet. since mass must also be balance, i.e $m_{1} = m_{2} + m_{3}$ $\\$

From the question the, the mass flow rate at the inlet $m_{1}}$ is 2Kg/s $\\$

Since 5% flow is delivered into the feedwater heating, $\\$

$m_{2} = 0.05m_{1} = 0.05 *2kg/s = 0.1kg/s$ $\\$

Also for the outlet 3 the remaining 95% will flow out. Hence

$m_{3} = 0.95m_{1} = 0.95 *2kg/s = 1.9kg/s$ $\\$

Now, from $m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$ we substitute values

$W_{out} = m_{1} h_{1}-m_{2} h_{2}-m_{3} h_{3}$

$W_{out} = (2kg/s)(3851.2KJ/Kg) - (0.1kg/s)(3206.5kJ/kg)- (1.9)(2682.4kJ/kg)$

$\\$

$W_{out} = 2285.19 kW$ .

Hence the power produced is 2285kW

7 0

3 years ago

A jetliner flying at an altitude of 10,000 m has a Mach number of 0.5. If the jetliner has to drop down to 1000 m but still main

andreev551 [17]

Answer

Assuming

At 10000 m height temperature T = -55 C = 218 K

At 1000 m height temperature T = 0 C = 273 K

$\dfrac{V_1}{C_1} =\dfrac{V_2}{C_2} = 0.5$

R = 287 J/kg K

$C_1 = \sqrt{\gamma RT_1} = \sqrt{1.4\times 287\times 218} = 295 m/s$

$C_2 = \sqrt{\gamma RT_2} = \sqrt{1.4\times 287\times 273} = 331 m/s$

$V_2 = \dfrac{V_1}{C_1}\timesC_2$

V₂ = V₁ ×1.1222

V₁ = 0.5 × C₁ = 0.5 × 295 = 147.5 m/s

V₂ = 1.1222 × 147.5 = 165.49 m/s

so, the jetliner need to increase speed by ( V₂ -V₁ )

= 165.49 - 147.5

= 17.5 m/s

6 0

4 years ago

Statement 1: All balls hit the ground at the same time. Statement 2: All balls hit the ground with the same force. Statement 3:

laiz [17]

Answer:

Statement 1: All balls hit the ground at the same time

Explanation:

When there is no resistance of air, the acceleration due to gravity experienced by all the bodies are same. So for falling bodies, neglecting the air resistance, the falling object will be weightless and therefore all the objects will hit the ground at the same time when there is nor air resistance and the objects are considered to be falling in vacuum.

7 0

3 years ago

A soil is at a void ratio e = 0.90 with a specific gravity of the solid particles Gs = 2.70.

Alexus [3.1K]

Answer:

The correct answers are:

a. % w = 33.3%

b. mass of water = 45g

Explanation:

First, let us define the parameters in the question:

void ratio e = $\frac{V_v}{V_s}$ = $\frac{\left\begin{array}{ccc}volume&of&void\end{array}\right}{\left\begin{array}{ccc}volume&of&solid\end{array}\right}------ (1)$

Specific gravity $G_{s}$ = $\frac{P_s}{P_w} =$ $\frac{\left\begin{array}{ccc}density&of&soil\end{array}\right}{\left\begin{array}{ccc}density&of&water\end{array}\right}------ (2)$

% Saturation S = $\frac{V_w}{Vv}$ × $\frac{100}{1}$ = $\frac{\left\begin{array}{ccc}volume&of&water\end{array}\right}{\left\begin{array}{ccc}volume&of&void\end{array}\right}$ × $\frac{100}{1}--------(3)$

water content w = $\frac{M_w}{M_s}$ = $\frac{\left\begin{array}{ccc}mass&of&water\end{array}\right}{\left\begin{array}{ccc}mass&of&solid\end{array}\right} ------(4)$