# 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

R_{1,2,3} = 2 + 4 + 2

R_{1,2,3} = 8 Ω

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

R_{5,6} = | 20•5 | |

25 |

R_{5,6} = 4 Ω

R_{4,5,6,7} = 4 + 4 + 4

R_{4,5,6,7} = 12 Ω

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

R_{8,9} = | 6•3 | |

9 |

R_{8,9} = 2 Ω

Now the circuit is as following.

The equivalent resistance of R_{1-3} and R_{4-7} is

R_{1-7} = | 8•12 | |

20 |

R_{1-7} = 4,8 Ω

Now the circuit is as following.

The equivalent resistance of three connected resistors in series.

R_{s} = 4.8 + 2 + 10

R_{s} = 16.8 Ω

The right answer is option C

**Solution 2**

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

R_{2,3,4} = | 9 | |

3 |

R_{2,3,4} = 3 Ω

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

R_{6,7} = | 6•3 | |

9 |

R_{6,7} = 2 Ω

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

R_{9,10} = | 10•15 | |

25 |

R_{9,10} = 6 Ω

Now the circuit is as following.

The equivalent resistance of seven resistors connected in series is

R_{1-11} = 6 + 3 + 1 + 2 + 4 + 6 + 4

R_{1-11} = 26 Ω

R_{s} = 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

R_{1,2} = | 12 | |

2 |

R_{1,2} = 6 Ω

The equivalent resistance of the branching-2 is

R_{5,6} = | 6•4 | |

6+4 |

R_{5,6} = 2.4 Ω

Now the circuit is as following.

The equivalent resistance of the resistors R_{4}, R_{5-6}, and R_{7} is

R_{4-7} = 7 + 2.4 + 2.6

R_{4-7} = 12 Ω

The resistor R_{3} and resistor R_{4-7} are parallel each other.

R_{3-7} = | 6•12 | |

18 |

R_{3-7} = 4 Ω

Now the circuit is as following.

The equivalent resistance of the circuit is

R_{s} = 6 + 10 + 6

R_{s} = 22 Ω

The current drawn from the circuit is

I_{s} = | 110 | |

22 |

I_{s} = 5 A

The current flowing through resistor R_{3} is I_{2} .

I_{2} = | I_{s}•R_{4-7} | |

R_{3} + R_{4-7} |

I_{2} = | 5•12 | |

18 |

I_{2} = 3.33 A

The voltage across R_{3} is

V_{R3} = I2•R3

V_{3} = 3.33•6

V_{3} = 20 V

The current I_{3} is

I_{3} = 5 -3.33

I_{3} = 1,67 A

The current flowing through resistor R_{5} is

I_{R5} = | I_{3}•6 | |

10 |

I_{R5} = | 1.67•6 | |

10 |

I_{R5} = 1 A

The voltage across resistor R5 is

V_{R5} = I_{R5}•R_{5}

V_{R5} = 1•4 = 4 V

V_{R5} + V_{R3} = 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

R_{1,2,3} = 5 + 3 + 2

R_{1,2,3} = 10 Ω

The equivalent resistance of the branching-2 is

R_{4,5,6} = 4 + 8 + 3

R_{4,5,6} = 15 Ω

The equivalent resistance of the branching-3 is

1 |
| |||||||||||||

R_{7,8,9} |

R_{7,8,9} = | 60 | |

30 |

R_{7,8,9} = 2 Ω

Now the circuit is as following.

The equivalent resistance of the two upper branches is

R_{1-6} = | 10•15 | |

25 |

R_{1-6} = 6 Ω

Now, R_{1-6}, R_{7-9} and R_{10} in series connected.

R_{s} = 6 + 2 + 2

R_{s} = 10 Ω

Now, let we find the main current.

I_{s} = | V | |

R_{s} |

I_{s} = | 60 | |

10 |

I_{s} = 6 A

The current flowing through the resistor R_{2} is I_{2}.

I_{2} = | Is•R_{4,5,6} | |

R_{4,5,}, + R_{1,2,3} |

I_{2} = | 6•15 | |

25 |

I_{2} = 3,6 A

I_{6} = I_{s} – I_{2}

I_{6} = 6 – 3.6

I_{6} = 2.4 A

The current I_{9}

Let we find the equivalent resistance of the resistors R_{7} and R_{8}.

R_{7,8} = | 4•12 | |

16 |

R_{7,8} = 3 Ω

I_{9} = | I_{s}•R_{7,8} | |

R_{7,8} + R_{9} |

I_{9} = | 6•3 | |

9 |

I_{9} = 2 A

I_{2} + I_{6} + I_{9} = 3,6 + 2,4 + 2

I_{2} + I_{6} + I_{9} = 8 A

The right answer is option C.

**Solution 5**

The equivalent resistance of the branching-1 is

R_{4,5} = | 4•6 | |

4+6 |

R_{4,5} = | 24 | |

10 |

R_{4,5} = 2.4 Ω

R_{3,4,5} = 2.8 + 2.4

R_{3,4,5} = 5.2 Ω

The equivalent resisance of the branching-2 is

R_{6,7} = | 12•8 | |

20 |

R_{6,7} = 4,8 Ω

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

R_{3-7} = 5.2 + 4.8

R_{3-7} = 10 Ω

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

R_{2-7} = | 15•10 | |

25 |

R_{2-7} = 6 Ω

Now, the R_{1}, R_{2-7} and R_{8} are in series connection.

R_{1-8} = 2 + 6 + 4

R_{1-8} = 12 Ω

The current drawn from the supply source is

I_{s} = | 36 | |

12 |

I_{s} = 3 A

Let we find the current I_{3}.

The R_{3-7} was 10 Ω, the R_{2} is 15 Ω.

I_{3} = | Is•R_{3-7} | |

R_{2} + R_{3-7} |

I_{3} = | 3•15 | |

25 |

I_{3} = 1.8 A

I_{4} = | 1.8•6 | |

10 |

I_{4} = 1.08 A

V_{4} = I4•R4

V_{4} = 1.08•4

V_{4} = 4.32 V

I_{7} = | 1,8•12 | |

20 |

I_{7} = 1.08 A

V_{7} = 1.08•8

V_{7} = 8.64 V

V_{4} + V_{7} = 8.64 + 4.32

= 12.96 V

The right answer is option B.

**Solution 6**

The equivalent resistance of these resistors,

1 |
| |||||||||||||

R_{s} |

1 |
| |||||

R_{s} |

R_{s} = 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

**Series Connecte Resistors Questions**

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