# Solutions of The Electric Circuits Test II

## Calculation the current and voltage in electric circuits. Calculation of the equivalent resistance in series-parallel circuits. Finding the current flowing through branches. Test II solutions.

Solution 1 If the current flowing through resistor R2 is 3 A, the current flowing through R3 is

 I2 = I1•R3 R2 + R3

 3 = I1•9 12

 I1 = 36 9

I1 = 4 A

The equivalent resistance of the branching-2 is

 R2,3 = 3•9 12

R2,3 = 2.25 Ohm

R1,2,3 = 5 + 2.25

R1,2,3 = 7.25 Ω

Voltage drop across branching -2 is

VR1-3 = R1,2,3•I1

VR1-3 = 7.25 • 4

VR1-3 = 29 V

The equivalent resistance of the branching-1 is

 R5,6 = 10 7

R5,6 = 1.43 Ω

R4,5,6,7 = 2 + 1.43 + 7

R4-7 = 10.43 Ω

The I4 current.

The voltage drop across branching -1 is equal to the voltage drop of the branching -2. Because of the voltages are equal in parallel arms.

VR4-7 = 29 V

 I4 = 29 10.43

I4 = 2.78 A

I2 + I4 = 4 + 2.78

I2 + I4 = 6.78 A

The sum of the I1 and I4 is 6.78 A. This current also flows through R8.

VR8 = (I2 + I4)•R8

VR8 = 6.78•8

VR8 = 54.24 V

The voltage of the supply source is

V = 29 + 54.24

V = 83.24 V

The right answer is option E

Solution 2 Let we find the current I3.

If VR3 = 10 V

 I3 = 10 5

I3 = 2 A

Because of the voltage across R3 is 10 V, the voltage across R4 is 10 V too.

 I4 = 10 10

I4 = 1 A

The current that flows through R5 is (I3 + I4)

VR5 = (I3 + I4)•R5

VR5 = 3•8 = 24 V

The sum of the voltages drop across in this arms is equal to the source voltage.

V = 10 + 24

V = 34 V

The right answer is option B

Solution 3 First, we must find the equivalent resistance.

The equivalent resistance of the branching -1 is

 R2,3,4 = 9 3

R2,3,4 = 3 Ω

R1-5 = 6 + 3 + 1

R1-5 = 10 Ω

The equivalent resistance of the branching -2 is

1
= 1
+ 1
 + 1 3
4
12
R7,8,9

 R7,8,9 = 12 8

R7,8,9 = 1.5 Ω

R6,7,8,9 = 8.5 + 1.5 Ω

R6,7,8,9 = 10 Ω

The R1-5 and R6-9 are parallel.

 R1-9 = 10 2

R1-9 = 5 Ω

The equivalent resistance of the circuit is

Rs = R1-9 + R10

Rs = 5 + 7

Rs = 12 Ω

The current flowing from the voltage source is

 Is = 48 12

Is = 4 A

Because of R1-5 and R6-9 are equal, the current flowing both arms are equal and 2 A.

 Ix = Is•R6-9 R1-5 + R6-9

 Ix = 4•10 20

Ix = 2 A

Iy = 4-2 = 2 A

The current flowing through resistor R3 is

 IR3 = 2•R3 R2+R3+R4

 IR3 = 2*9 27

IR3 = 0.667 A

V3 = 0.667•9

V3 = 6 V

Let we find the equivalent resistance of the resistors R7 and R9 for find the current that flows through R3.

 R7,9 = 36 15

R7-9 = 2.4 Ω

Now, we can find the current I8.

 IR8 = 2•2.4 6.4

IR8 = 0.75 A

V8 = I8•R8

V8 = 0.75•4

V8 = 3 V

V3 + V8 = 6 + 3 = 9 V

The right answer is option C

Solution 4 First, Let we find the equivalent resistance of the upper arm (branch 1).

There are resistors R3 – R8 in the upper arm. The resistors R3 and R4 are parallel amongst themselves, the resistors R5 and R6 are parallel amongst themselves too.

 R3,4 = 6•12 18

R3,4 = 4 Ω

 R4,5 = 8 = 4 Ω 2

R3-8 = 4 + 4 + 2 + 2

R3-8 = 12 Ω

The equivalent resistance of the branching -2 is

R1,2 = 3 + 5

R1,2 = 8 Ω

Branch -1 and branch-2 are parallel.

 Rs = R3-8 • R1-2 R3-8 + R1-2

 Rs = 12•8 20

Rs = 4.8 Ω

The main current is

 Is = 96 4.8

Is = 20 A

The current flowing through branching -1 is

 Ix = Is•R1,2 R1,2 + R3-8

 Ix = 20•8 20

Ix = 8 A

The current I3 flowing through resistor R3 is

 I3 = Ix•R4 R3 + R4

 I3 = 8•12 18

I3 = 5.33 A

The right answer is option A

Solution 5

The equivalent resistance of the circuit is,

Rs = R1 + R2 + R3 + R4

Rs = 6 + 3 + 5 + 1

Rs = 15 Ω

The current drawn from the supply source is

 Is = 60 = 4 A 15

Suppose the resistor that we add to the circuit is R5 and R6. The equivalent resistance of these resistors is

 R5,6 = 10 2

R5,6 = 5 Ω

The resistor R5 will be connected in series to the circuit.

Now the equivalent resistance will be

Rs = 15 + 5

Rs = 20 Ω

In this case, the current drawn from the supply source is

 Is = 60 20

Is = 3 A

The right answer is option E

Solution 6

We suppose that

R1 = 12 Ω

R2 = 15 Ω

R3 = 10 Ω

R4 = 3 Ω

R5 = 2 Ω

R1, R2 and R3 are parallel. The resistors R3 and R4 will connect to these resistors.

The equivalent resistance of the R1, R2 and R3

1
= 1
+ 1
 + 1 10
15
12
R1,2,3

 R1,2,3 = 60 15

R1,2,3 = 4 Ω

If it is connected two resistors R4 = 3 and R5 = 2 in series to the circuit, the equivalent resistance of the circuit is

R1-5 = 4 + 3 + 2

R1-5 = 9 Ω

The current drawn from the source is

 Is = 36 9

Is = 4 A

The right answer is option B

The Questions of This Test

Electric Circuit Questions 1

RISE KNOWLEDGE

November 9 2018

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