Ask question

Ask question

All categories

3 years ago

12

As more resistors are added in series across a constant voltage source, the power supplied by the source: A) does not change. B)

increases. C) decreases. D) is always zero

1 answer:

Anon25 [30]3 years ago

4 0

As series resistors are added, the resistance is added directly so the total resistance will be equal to

$R_T = R_1+R_2+R_3... R_{\infty}$

Since the power is determined as

$P = \frac{V^2}{R}$

We have there that

$P \propto \frac{1}{R}$

The power is inversely proportional to the increase in resistance, so it will tend to decrease as more resistors are added in series.

The correct answer is: C.

You might be interested in

Question 3 Please, i need help! Thank you

Tom [10]

I don't know this 1 I'm sorry I can't help you

4 0

3 years ago

An electric furnace is to melt 40 kg of aluminium/hour. The initial temperature of aluminium is 32°C. Given that aluminium has s

gizmo_the_mogwai [7]

Answer:

Part a)

P = 13.93 kW

Part b)

R = 8357.6 Cents

Explanation:

Part A)

heat required to melt the aluminium is given by

$Q = ms\Delta T + mL$

here we have

$Q = 40(950)(680 - 32) + 40(450 \times 10^3)$

$Q = 24624 kJ + 18000 kJ$

$Q = 42624 kJ$

Since this is the amount of aluminium per hour

so power required to melt is given by

$P = \frac{Q}{t}$

$P = \frac{42624}{3600} kW$

$P = 11.84 kW$

Since the efficiency is 85% so actual power required will be

$P = \frac{11.84}{0.85} = 13.93 kW$

Part B)

Total energy consumed by the furnace for 30 hours

$Energy = power \times time$

$Energy = 13.93 kW\times 30 h$

$Energy = 417.9 kWh$

now the total cost of energy consumption is given as

$R = P \times 20 \frac{Cents}{kWh}$

$R = 417.9 kWh\times 20 \frac{cents}{kWh}$

$R = 8357.6 Cents$

3 0

3 years ago

Air rushing against an airplane is an example of _____friction.

slava [35]

The word "static" would be known to be friction as air rushing against an airplane

6 0

3 years ago

What is the efficiency of the<br> machine if the input is 263 J and<br> the output is 142 J?

Papessa [141]

Answer: 57.79%

Explanation: 152J/263J=.577946768 or 57.79% or roundedthe nearest whole percent is 58%

7 0

3 years ago

A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of

GREYUIT [131]

Answer:

$a=190\ m/s^2$

Explanation:

Mass of a hockey puck, m = 0.17 kg

Force exerted by the hockey puck, F' = 35 N

The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

$a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2$

So, the acceleration of the hockey puck is $190\ m/s^2$ .

5 0

3 years ago