# Solutions Of The Electric Circuits Test I

## Finding equivalent resistance in circuits that consist of series and parallel combinations. Calculation the current flow in the arms of series-parallel circuits. Solutions of the test-1

Solution 1 The equivalent resistance of the resistors that on the branching-1 is

R1,2,3 = 2 + 4 + 2

R1,2,3 = 8 Ω

The equivalent resistance of the resistors that on the branch-2 is

 R5,6 = 20•5 25

R5,6 = 4 Ω

R4,5,6,7 = 4 + 4 + 4

R4,5,6,7 = 12 Ω

The equivalent resistance of the resistors that on the branching-3 is

 R8,9 = 6•3 9

R8,9 = 2 Ω

Now the circuit is as following. The equivalent resistance of R1-3 and R4-7 is

 R1-7 = 8•12 20

R1-7 = 4,8 Ω

Now the circuit is as following. The equivalent resistance of three connected resistors in series.

Rs = 4.8 + 2 + 10

Rs = 16.8 Ω

The right answer is option C

Solution 2 The equivalent resistance of the resistors that on the branching-1 is

 R2,3,4 = 9 3

R2,3,4 = 3 Ω

The equivalent resistance of the resistors that on the branching-2 is

 R6,7 = 6•3 9

R6,7 = 2 Ω

The equivalent resistance of the resistors that on the branching-3 is

 R9,10 = 10•15 25

R9,10 = 6 Ω

Now the circuit is as following. The equivalent resistance of seven resistors connected in series is

R1-11 = 6 + 3 + 1 + 2 + 4 + 6 + 4

R1-11 = 26 Ω

Rs = 26 Ω

The right answer is option B

Solution 3 Firstly, we must find the equivalent resistance of the circuit.

The equivalent resistance of the branching-1 is

 R1,2 = 12 2

R1,2 = 6 Ω

The equivalent resistance of the branching-2 is

 R5,6 = 6•4 6+4

R5,6 = 2.4 Ω

Now the circuit is as following. The equivalent resistance of the resistors R4, R5-6, and R7 is

R4-7 = 7 + 2.4 + 2.6

R4-7 = 12 Ω

The resistor R3 and resistor R4-7 are parallel each other.

 R3-7 = 6•12 18

R3-7 = 4 Ω

Now the circuit is as following. The equivalent resistance of the circuit is

Rs = 6 + 10 + 6

Rs = 22 Ω

The current drawn from the circuit is

 Is = 110 22

Is = 5 A

The current flowing through resistor R3 is I2 .

 I2 = Is•R4-7 R3 + R4-7

 I2 = 5•12 18

I2 = 3.33 A

The voltage across R3 is

VR3 = I2•R3

V3 = 3.33•6

V3 = 20 V

The current I3 is

I3 = 5 -3.33

I3 = 1,67 A

The current flowing through resistor R5 is

 IR5 = I3•6 10

 IR5 = 1.67•6 10

IR5 = 1 A

The voltage across resistor R5 is

VR5 = IR5•R5

VR5 = 1•4 = 4 V

VR5 + VR3 = 20 + 4 = 24 V

The right answer is option E

Solution 4 Let we find firstly the equivalent resistance of the circuit.

The equivalent resistance of the branching-1 is

R1,2,3 = 5 + 3 + 2

R1,2,3 = 10 Ω

The equivalent resistance of the branching-2 is

R4,5,6 = 4 + 8 + 3

R4,5,6 = 15 Ω

The equivalent resistance of the branching-3 is

1
=1
+1
 + 1 16
12
4
R7,8,9

 R7,8,9 = 60 30

R7,8,9 = 2 Ω

Now the circuit is as following. The equivalent resistance of the two upper branches is

 R1-6 = 10•15 25

R1-6 = 6 Ω

Now, R1-6, R7-9 and R10 in series connected.

Rs = 6 + 2 + 2

Rs = 10 Ω

Now, let we find the main current.

 Is = V Rs

 Is = 60 10

Is = 6 A

The current flowing through the resistor R2 is I2.

 I2 = Is•R4,5,6 R4,5,, + R1,2,3

 I2 = 6•15 25

I2 = 3,6 A

I6 = Is – I2

I6 = 6 – 3.6

I6 = 2.4 A

The current I9

Let we find the equivalent resistance of the resistors R7 and R8.

 R7,8 = 4•12 16

R7,8 = 3 Ω

 I9 = Is•R7,8 R7,8 + R9

 I9 = 6•3 9

I9 = 2 A

I2 + I6 + I9 = 3,6 + 2,4 + 2

I2 + I6 + I9 = 8 A

The right answer is option C.

Solution 5 The equivalent resistance of the branching-1 is

 R4,5 = 4•6 4+6

 R4,5 = 24 10

R4,5 = 2.4 Ω

R3,4,5 = 2.8 + 2.4

R3,4,5 = 5.2 Ω

The equivalent resisance of the branching-2 is

 R6,7 = 12•8 20

R6,7 = 4,8 Ω

The equivalent resistance of the branch-1 and branch-2 is

R3-7 = 5.2 + 4.8

R3-7 = 10 Ω

The branching-3 is paralel the branch-1 + branch-2

 R2-7 = 15•10 25

R2-7 = 6 Ω

Now, the R1, R2-7 and R8 are in series connection.

R1-8 = 2 + 6 + 4

R1-8 = 12 Ω

The current drawn from the supply source is

 Is = 36 12

Is = 3 A

Let we find the current I3.

The R3-7 was 10 Ω, the R2 is 15 Ω.

 I3 = Is•R3-7 R2 + R3-7

 I3 = 3•15 25

I3 = 1.8 A

 I4 = 1.8•6 10

I4 = 1.08 A

V4 = I4•R4

V4 = 1.08•4

V4 = 4.32 V

 I7 = 1,8•12 20

I7 = 1.08 A

V7 = 1.08•8

V7 = 8.64 V

V4 + V7 = 8.64 + 4.32

= 12.96 V

The right answer is option B.

Solution 6

The equivalent resistance of these resistors,

1
= 1
+ 1
 + 1 10
60
20
Rs

1
 = 10 60
Rs

Rs = 6 Ω

If a resistor R is connected in series to the equivalent resistance Rs, equivalent resistance of the circuit will be 8 Ohm.

8 = 6 + R

R = 2 Ω

It must be 2Ω.

The right answer is option A

Questions of This Test

Series Connecte Resistors Questions

Electric Circuit Questions II

RISE KNOWLEDGE

November 5 2018

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