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2 years ago

As resistors are added in series to a circuit, the total resistance will

Engineering

1 answer:

olganol [36]2 years ago

8 0

The equivalent of the resistance connected in the series will be Req=R₁+R₂+R₃.

<h3>
What is resistance?</h3>

Resistance is the obstruction offered whenever the current is flowing through the circuit.

So the equivalent resistance is when three resistances are connected in series. When the resistances are connected in series then the voltage is different and the current remain same in each resistance.

V eq = V₁ + V₂ + V₃

IReq = IR₁ + IR₂ + IR₃

Req = R₁ + R₂ + R₃

Therefore the equivalent of the resistance connected in the series will be Req=R₁+R₂+R₃.

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

a) To calculate the lower and upper limits of water content:

when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.

when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.

Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.

To get the relationship between water content and saturation, we will manipulate the equations above;

w = $\frac{M_w}{Ms}$

Recall; mass = Density × volume

w = $\frac{V_wP_w}{V_sP_s} ------(5)$

From eqn. (2) $G_{s}$ = $\frac{P_s}{P_w}$

∴ $\frac{1}{G_s} = \frac{P_w}{P_s} ------(6)$

putting eqn. (6) into (5)

w = $\frac{V_w}{V_sG_s} -----(7)$

Again, from eqn (1)

$V_s = \frac{V_v}{e}$

substituting into eqn. (7)

$w = \frac{V_w}{\frac{V_v}{e}{G_s} } = \frac{V_w e}{V_vG_s} \\ but \frac{V_w}{V_v} = S$

∴ $w = \frac{Se}{G_s} -----(8)$

With eqn. (7), we can calculate

upper limit of water content

when S = 100% = 1

Given, $G_{s} = 2.7, e= 0.9$

∴ $w= \frac{0.9*1}{2.7} = 0.333$

∴ %w = 33.3%

Lower limit of water content

when S = 1% = 0.01

$w= \frac{0.01*0.9 }{2.7} = 0.0033$

∴ % w = 0.33%

b) Calculating mass of water in 100 cm³ sample of soil ( $P_w=\frac{1_g}{cm^{3} }$ )

Given, $V_{s} = 100 cm^{3 }$ , S = 50% = 0.5

%S = $\frac{V_w}{V_v}$ × $\frac{100}{1}$ = $\frac{V_w}{eV_s}$ × $\frac{100}{1}$

0.50 = $\frac{V_w}{0.9* 100} = 45cm^{3}$

mass of water = $P_wV_w= 1 * 45 = 45_{g}$

7 0

3 years ago

A Carnot heat pump maintains the temperature of a house at 22 C in the winter. On a day when the average outdoor temperature is

wlad13 [49]

Answer:

There are 3 sub answers from the above question.

1. the time the heat pump ran that day = 89,495.7726 ÷ 4.8 = 18,645 seconds or 310.75 minutes or 5.18 hours

2.Cost= (4.8 x 5.18) x 11 = 273.5 g (We assume in the question that the "g" signifies currency).

3. New cost = 4,033.33 g (We assume in the question that the "g" signifies currency).

Explanation:

1. To find how long the heat pump ran on that day, we have to look at the coefficient of performance of the pump. This will then enable us find the work done by the heat pump.

Coefficient of performance of the heat pump, C $H$ = 1 ÷ ( (1 - (TL ÷ TH) )

Heated space, TH = 22°C = 295k

Out door air temperature, TL = 2 °C = 275k

where, 1°C = 273k

= 1 ÷ ( (1 - ( 275k ÷ 295k) )

C $H$ = 14.7493

Then the work done by the heat pump

W = QH ÷ CH

Where, the rate of heat loses, Q $H$ given = 55,000 kJ/h

= (55,000 x 24) ÷ 14.7493 = 89,495.7726 kJ

Where, 24 is the number of hours in a day

Therefore, the time the heat pump ran that day = the work done by the heat pump ÷ the power the heat pump consumes during operation

the time the heat pump ran that day = 89,495.7726 ÷ 4.8 = 18,645 seconds or 310.75 minutes or 5.18 hours

2. To find the total heating costs assuming an average price of electricity of 11 g/kWh.

In the question "g" signifies currency.

Therefore,

cost = (the power the heat pump consumes during operation x the time in hours the heat pump ran that day) x price in g

Cost= (4.8 x 5.18) x 11 = 273.5 g

3. To find the heating cost for the same day if resistance heating is used instead of a heat pump.

If the resistance heating was used, the cost will be:

New cost = the rate of heat loses x price in g

New cost = ( (55,000 x 24) ÷ 3600) ) x 11

New cost = 4,033.33 g

Where, 24= hours in a day

and, 3600 = seconds in one hour

3 0

3 years ago