Energy In Electric Circuits Test-II Solutions

Calculations of the power and energy in series and parallel circuits. Calculating the power and energy in circuits created with bulb, motor and resistors. Solutions of the test-II questions.


Solutions 1

Let we find the current flowing through in electrical motor.

V = 380 V

R = 95 Ω

I = 380
95




I = 4 A

P = V*I

P = 380*4

P = 1520

1 hp was 746 Watt. Let we establish proportion.


1 hp            746 W

x                1520


x = 1520
746



x = 2.04 hp


The right answer is option E


Solution 2

First, let we calculate the heat energy given to the water by container K and L

For this, we must find the current flowing through each container.

The sum of the resistance values of the two container is 

Rs = 18 + 26 = 44 Ω

Current of the circuit is

Is =110
44



Is = 2,5 A

Power of the container K

PK = (2.5)2 *18

PK = 6,25*18

PK = 112.5 W


Power of the container L

PL = (2.5)2*26

PL = 162.5 W


20 minutes = 20*60 = 1200 seconds

Given energy to the water by the container K in 1200 seconds is

WK = 112.5*1200

WK = 135000 J


Given energy to the water by the container L in 1200 seconds is


WL = 162.5*1200

WL = 195000 J


Now, let we calculate last temperatures of the waters in the containers.

QK = m*c*t

135000 = 600*4*(t2 – 10)

1350 = 24(t2 – 10)

56.25 = t2 – 10

t2 = 66.25 °C

T1 = 66.25 °C


The last temperature of water in the container K is 66.25 °C

Let we calculate last temperature of the water in the container L too.

QL = m*c*Dt

195000 = 600*4*(t2 – 10)

1950 = 24(t2 – 10)

81.25 = t2 – 10

t2 = 91.25 °C

T2 = 91.25 °C


The right answer is option C


Solution 3

The voltage of all three bulb are equal and 100 V.

Let we find the power of each lamp.


PL1 = V2
RL1



PL1 = 1102
160



PL1 = 75.6 W

PL2 = 1102
200



PL2 = 60.5 W

PL3 = 1102
220




PL3 = 55 W

Total power of all three bulb is

Ptotal = 75.6 + 60.5 + 55

Ptotal = 191.1 W

1 hour = 3600 seconds

Wtotal = 191.1*3600

Wtotal = 687960 J

Wtotal = 687.96 kJ

The right answer is option D


Solution 4

Let we find the power of the bulbs.

PL1 = 1102
220



PL1 = 55 W

PL2 = 1102
100



PL2 = 121 W

If the bulb L1 operates in 8 hours per a day, the amount of energy to be consumed in one day is

Wday = 55*8

Wday = 440 Wh = 0.44 kWh

The amount of energy to be consumed in 30 day is

Wmonth = 0.44*30

Wmonth = 13,2 kWh

1 kWh = 0.2 USD,


1 kWh               0,2 USD

13,2 kWh            x


x = 2,64 USD

Cost L1 = 2.64 USD

If the bulb L2 operates in 8 hours per a day, the amount of the energy to be consumed in one day is

Wday = 121*8

Wday = 968 W


The amount of the energy to be consumed in 30 day is

Wmonth = 968*30

Wmonth = 29040

Wmonth = 29.04 kWh


1 kWh                 0.2 USD

29.04 kWh              y


y = 5.80 USD


y – x = 5.8 – 2.64

y – x = 3.16 USD


The right answer is option B


Solution 5

P = 5 hp

1 hp = 746 W, power of the appliance is

Wc = 746*5

Wc = 3730 W


P = V*I

3730 = 380*I

I = 9.8 A


The right answer is option A


Solution 6

Elcenergy_t2s6


First, we finds the equivalent resistance of the circuit. Then we find the main current (Is) of the circuit.

We can find the currents I1 and I5 using the current Is.

Using the equailities of P = I2*R and E = P*t we can find the value of E1 and E5.

R3,4 = 6 Ω

R1,3,4 = 6
2



R1,3,4 = 3 Ω

R2,5 = 12*4
16



R2,5 = 3 Ω

Rs = R1,3,4 + R2,5 + R6

Rs = 3 + 3 + 4

Rs = 10 Ω

Is = 60
10



Is = 6 A

I1 = Is*R3,4
R3,4 + R1




I1 = 6*6 = 3A
12



PR1 = 32*6

PR1 = 54 W

E1 = 54*t

I5 =6*4
16



I5 = 1.5 A

PR5 = (1.5)2*12

PR5 = 27 W

E5 = 27*t

E1 
= 54*t
27*4
E5




= 2


The right answer ip option C


Solution 7

Let we calculate the energy consumed in one day in the house.

Washing machine→ 2000 W*1 hour = 2000 Wh

W.M. = 2 kWh

Television → 30 W*5 hour = 150 Wh

TV = 0.15 kWh


Bulbs → 2*75 W*6 hours = 900 Wh

Bulbs = 0.9 kWh


Refrigerator → 100 W*8 hours = 800 Wh

Refrigerator = 0.8 kWh


The total energy consumed in one day.

Tday = 2 + 0.15 + 0.9 + 0.8

Tday = 3.85 kWh

The total energy consumed in thirty day.

Tmonth = 3.85*30

Tmonth = 115.5 kWh


Total cost = 115.5*0.2

Total cost = 23.1 USD


The right answer is option E


Solution 8

The equivalent resistance of the circuit is

R1,2 = 6 + 4 = 10 Ω

R3,4 = 3 + 12 = 15 Ω

R1-4 =10*15
25



= 6 Ω

Rs = R1-4 + R5 = 6 + 4 = 10 Ω


Total power is

P =V2
R



P =602
10



P = 360 W

Consumed energy is

W = P*t

15 minutes =15= 0.25 hours
60




W = 360*0.25

W = 90 kWh



The right answer is option B


Electrical Energy Test-2 Questions

Electrical Energy Test-1

Power Questions In Electric Circuits



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 December 15 2018

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