Solutions Of The Electric Power Test II

Subject of electrical circuits. The consumed power in serial and parallel circuits. Total power taken from voltage source. Solutions of the solved questions II.


Solution 1

Power_t2sls1


If we know the power in R6, we can find the current I6 using this power.

If we can find the current I6, we can find the current I3 too.

P6 = 16 W

P6 = (I6)2*R6

16 = (I6)2*4

(I6)2 = 4

I6 = 2 A

I6 = I3*12
12+6



2 = I3*12
18




36 = 12*I3

I3  = 3 A


The right answer is option B


Solution 2

Power_t2sls2


We can find the current flow R3 and the voltage drop across R3 using the power in the R3.

If we can find the current Ix that flows in this arm, we can find the current Iy.

If we can find the equivalent resistance of circuit and total current of the circuit, we can calculate the voltage of the supply source.

P3 = 24 W

24 = (I3)2*R6

24 = (I3)2*6

I3 = 2 A

VR3 = 2*6

VR3 = 12 V

The voltage drops in parallel arms are equal.

If resistance and voltage in the arms are known, the current flowing through these arms can be find.


VR4 = VR2 = VR3 

VR4 = 12 V

I4 = 12
12



I4 = 1 A

VR2 = 12 V

I2 = 12
4



I2 = 3 A

Total current of lower arm is

Ix = I2 + I3 + I4

Ix = 3 + 2 + 1

Ix = 6 A

The voltage drop across R1 is

VR1 = Ix*R1

VR1 = 6*6

VR1 = 36 V


The voltage drop in the lower arm is equal to the sum of the VR1 with voltage that one of the parallel arms.

Vbottom = VR1 + VR4

Vbottom = 12 + 36 = 48 V

Because of the voltage drops across parallel arms are equal, the voltage of the upper arm is 48 V too.

Let’s find the equivalent resistance of the upper arm.

R6,7 =20*5
25



R6,7 = 4 Ω

R5-7 = 4 + 4 = 8 Ω

Iy = 48
8



Iy = 6 A

The current I8 is equal the sum of the currents Ix and Iy.


I8 = Ix + Iy = 6 + 6

I8 = 12 A

VR8 = 12*6 = 72 V


Total voltage = Vlower + VR8

Vsource = 48 + 72

Vsource = 120 V


The right answer is option E


Solution 3

Power_t2sls3


The equivalent resistance of the Branching -2 is

R7,8 = 6*4
10



R7,8 = 2,4 Ω

RB2 = R6 + R7,8 + R9

RB2 = 2 + 2,4 + 1,6

RB2 = 6 Ω

The equivalent resistance of the Branching -1 is

R4,5 =12*8
20



R4,5 = 4,8 Ω

RB1 = R3 + R4,5 

RB1 = 7,2 + 4,8 = 12 Ω


Branch -1 and Branch-2 are parallel. The equivalent resistance of these branches is,

RB1-2 = 12*6
18



RB1-2 = 4 Ω


Total resistance of the circuit is

Rs = R1 + R2 + RB1-2

Rs  = 3 + 5 + 4

Rs = 12 Ω


Main current of the circuit is

Is =48
12



Is = 4 A

Ix = Is = 4 A

Iz = Ix*RB1
RB1 + RB2




Iz = 4*12
18



Iz = 2,67 A

I8 = Iz*R7
R7+R8




I8 = 2,67*6
10




I8 = 1,6 A


P8 = (I8)2*R8

P8 = 2,56*4

P8 = 10,24 W


The right answer is option C


Solution 4

Power_t2sls4


First we must find the equivalent resistance of the circuit. Then we find main current of the circuit. We can find the power drawn from the source by equality of the V*I or I2R.

R3-5 = 30 Ω

R1-2 = 30

R1-5 =30 = 15 Ω
2




R1-8 = 15 + 2 + 5 + 8

R1-8 = 30 Ω

Rs = 30 Ω

Is =60
30



Is = 2 A

P = V*I

P = 60*2 

P = 120 W 

The right answer is option A


Solution 5

If three resistors of 16 ohms are connected in parallel to a 48 V source, the voltage drop at each resistor is 48 V.

Current flowing through each resistor is

I1 = I2 = I3

I1 = 48
16



I1 = 3 A

I2 = 3 A

I3 = 3 A

Total current drawn from source is

I1 + I2 + I3 = 9 A 

Power in each resistors is

P1 = (I1)2*R1

P1 = 9*16 = 144 W

If three resistors are parallel and equal to each other, total power drawn from source is found by multipliying of the power of one of resistor by 3.

Pt = 144*3 = 432 W

If an 8 Ohm resistor is connected to these resistors in parallel, the voltage drop across this resistor is be 48 V too. Suppose the current flowing through this resistor is I4.

I4 =48
8



I4 = 6 A 


In this case, there is no any change in the current and power of the 3 resistor of 16 Ohm . Only, due to fourth resistance an additional current and power is transmitted to the circuit.

Power in the resistor of 8 ohm is


P4 = 36*8 = 288 W

The total power will increase 288 W.


The right answer is option B


Solution 6

The equivalent resistance of three resistors of 8,12 and 15 Ohm is

Rs = 18 + 12 + 15

Rs = 45 Ω


In this case, current drawn from the source is

Is =90
45



Is = 2 A

In this case, total power drawn from circuit is

Pt = V*I

Pt = 90*2 = 180 W


In addition, if a resistor of 15 Ohm is connected in series to these resistors, the equivalent resistance of the circuit is

Rs = 45 + 15

Rs = 60 Ω

In this case, the total current is

Is =90
60



Is = 1,5 A


In last case, total power drawn from source is

Pt = V*I

Pt = 90*1,5

Pt = 135 W


Difference = 180 – 135

Difference = 45 W


The power drawn from source has been decreased 45 W.

The right answer is option C


Questions of This Test

Power Questions In Electric Circuits Test-1




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November 28 2018

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