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

Circuit_question_t2s1i1


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.

Circuit_question_t2s1i1b


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.

 Circuit_question_t2s1i1c


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

Circuit_question_t2s2i2


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.


Circuit_question_t2s2i2b 


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

Circuit_question_t2s3i1


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.

Circuit_question_t2s3i1b2


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.

Circuit_question_t2s3i1c


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

Circuit_question_t2s4i1


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.

Circuit_question_t2s4i1b


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

Circuit_question_t2s5i1


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

  • WRITE COMMENT
  • NAME SURNAME(or nick)
  • COMMENT
COPYRIGHT© ALL RIGHTS RESERVED