Energy Test-1 Solutions

Subject of electric circuits. Calculation of consumed energy in serial and parallel circuits. Calculation of  energy that consumed in any device. Solutions of the test-1 questions.


Solutions 1

Electric_energy_t1s1


First we finds the equivalent resistance of the circuit. We find the main current using this equivalent resistance.

If we find the main current, we can find the voltages, Powers, and current of all resistors.


The resistors R5 and R6 are series connected.

R5,6 = 6 + 4 = 10 Ω

The resistors R4 and R5,6 are parallel.

R4-6 = 15*10
25



R4-6 = 6 Ω

The resistors R3 and R4-6 are series connected.

R3-6 = 9 + 6 = 15 Ω


The resistors R2 and R3-6 are parallel

R2-6 = 15*10
25




R2-6 = 6 Ω

The resistors R1 and R2-6  are series connected.

R1-6 = 14 + 6 

R1-6 = 20 Ω

The equivalent resistance of the circuit is

Rs = 20 Ω


The main current is

Is =110
20




Is = 5.5 A


I1 = Is = 5 A

P1 = (I1)2 * R1

P1 = 30.25*14

P1 = 423.5 W

I3 = Is*R2
R2 + R3-6




I3 = 5.5*10
25



I3 = 2.2 A

I5 = Is*R4
R4 + R5,6




I5 = 2.2*15
25



I5 = 1.32 A

P6 = (I5)2*R6

P6 = 1.74*4

P6 = 6.96


P1 + P6 = 423.5 + 6.96

= 430.46 W

The energy lost in 1 second in the R1 and R2 is 

430.46*1 = 430.46 J


The correct answer is option B


Solution – 2

Electric_energy_t1s2


First, let we find the equivalent resistance of the circuit.


R4,5 = 12*4
16



R4,5 = 48
16




R4,5 = 3 Ω

R3-5 = 7 + 3 = 10 Ω

R1,2 = 6 + 4 = 10 Ω


The resistor R1,2 and resistor R3-5 are parallel.

R1-5 = 10
2




R1-5 = 5 Ω

R1-6 = 5 + 9 = 14 Ω


Main current of the circuit is,

Is = 42
14



Is = 3 A


I1 = Is*R3-5
R1,2 + R3-5




I1 = 3*10
20



I1 = 1,5 A

P2 = (1,5)2*R2

P2 = 2.25*4

P2 = 9 W

10 minutes = 60*10 = 600 seconds

W2 = 9*600

W2 = 5400 J


The correct answer is option A.


Solution 3

The energy consumed by any device is found as follows.

W = P.t

Here, "t" is given in seconds.

1,5 hours = 1,5*60 = 90 minutes

90 minutes = 90*60 = 5400 seconds


W = 80*5400

W = 432000 j

W = 432 kJ


The correct answer is option C.


Solution – 4 

The formula that gives the power of the lamp is

P = V*I

P = I2*R

P = 50, V = 110 V için,

50 = 110*I

I = 0.45 A

50 = (0.45)2*R

50 = 0.2*R

R = 250 Ω


The correct answer is option E


Solution 5

The consumed electric energy in "t" seconds in any electrical appliance is found as follows.

W = P.t 

3 hours = 3*60 = 180 minutes

180 minutes = 180 * 60 = 10800 seconds

2700000 J energy is consumed in 10800 seconds.



270000 = P*10800

2700 = 108*P

P = 25

P = I2*R

25 = I2*100

I2 = 0.25

I = 0.5 A


The correct answer is option B.


Solution 6

Electric_energy_t1s6


W = P.t

9000 = P*1800

P = 5 W

P = I2*R1

5 = I2*5

I = 1 A

I1 = 1 A

Equivalent resistance of the resistors in lower arm.

R1,2 = 5 + 7 = 12 Ω

Equivalent resistance of the resistors in upper arm.

R3,4,5 = 3 + 2 + 1

R3,4,5 = 6 Ω

The current I1 = 1 A, let we find the current I2.


To find the current I2, we applies the rule which the voltages are equal in parallel arms.

If the current flowing through in resistor R is 1 A, this mean the current 1 A flowing through in lower arm.

The voltage of the lower arm is

VR1,2 = 1*12 = 12 V

The voltage in upper arm must be is 12 V too.

VR3-5 = 6*I2

12 = 6*I2

I2 = 2 A

The current I3 is equal to the sum of the currents I2 and I3.

I3 = I1 + I2 = 3 A

R6,7,8 = 16 Ω

VR6-8 = 3*16 = 48 V

The total voltage is

V = VR1,2 + VR6-8 

= 12 + 48

V = 60 V


The correct answer is option D


Solution 7

Electric_energy_t1s7


Let we find the amount of heat increasing the temperature of the water from 10 degrees to 90 degrees.

Q = m.c.t

Q = 250*1*(90 – 10)

Q = 250*80

Q = 20000 Cal


Let we find 20000 Cal how much Joules.


1 cal                  4 J

20000 cal               x


x = 80000 J


The resistor has given 80000 J heat energy in 3 minutes.

3 minutes = 180 seconds

W = P*t

80000 = 180*P

P = 444.4 W

P = V*I

444.4 = 60*I

I = 7.4 A


V = I*R

60 = 7,4*R

R = 8,1 Ohm


The right answer is option B


Solution 8

1 Watt*second = 1 Joule

1 hours = 60 minutes = 3600 seconds

1Watt*3600 seconds = 3600 J

3600 second = 1 hour

1Watt*hour = 3600 J

1 kW = 1000 W

1kWh = 1000W*hour

1000Watt*hour = 3600*1000

1kWh = 3600000 J

1 kWh = 3600 kJ

178kWh = 178*3600 kJ

= 640800 kJ


The correct answer is option C


Solution 9

If a 1500 Watt device is operated for 3 hours a day, its amount of consumed energy in one day is

1500 Watt*3 = 4500 Watt*saat 

4500 Watt*hour = 4,5 kWh

1 kWh 0,2 USD 

4,5 kWh x

X = 9 USD

The total daily amount is 9 USD. 30 days electricity cost of this device is


30*9 = 270 USD 


The correct answer is option A


Solution 10

We calculate the amount of the energy that will increase the temperature of the room by 15 degrees.

W = 15*240

W = 3600 J


We calculate the heat energy that a 1000 watt heater gives in 1 second.

1000.1 = 1000 J 

1000 J = 1 kJ

The heater dissipates 1 k J energy in 1 second.

Let we find how long must operated the heater for increase the temperature of the room by 1 degrees.


240 = 1 kJ*t

t = 240 seconds

240 seconds = 4 minutes


The heater must operated during 4 minutes for increase the temperature of the room by 1 degrees.

To increase by 15 °C 

15*4 = 60 

It must operated during 60 minutes

60 minutes is equal to the 1 hour. If the heater is operated during 1 hour, it's consumption cost is 0.2 USD.

The right answer is option D


Electric Energy Questions-1

Electric Energy Subject Expression



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December 7 2018

 

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