# 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 R_{2} is 3 A, the current flowing through R_{3} is

I_{2} = | I_{1}•R_{3} | |

R_{2} + R_{3} |

3 = | I_{1}•9 | |

12 |

I_{1} = | 36 | |

9 |

I_{1} = 4 A

The equivalent resistance of the branching-2 is

R_{2,3} = | 3•9 | |

12 |

R_{2,3} = 2.25 Ohm

R_{1,2,3} = 5 + 2.25

R_{1,2,3} = 7.25 Ω

Voltage drop across branching -2 is

V_{R1-3} = R_{1,2,3}•I_{1}

V_{R1-3} = 7.25 • 4

V_{R1-3} = 29 V

The equivalent resistance of the branching-1 is

R_{5,6} = | 10 | |

7 |

R_{5,6} = 1.43 Ω

R_{4,5,6,7} = 2 + 1.43 + 7

R4-7 = 10.43 Ω

The I_{4} 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.

V_{R4-7} = 29 V

I_{4} = | 29 | |

10.43 |

I_{4} = 2.78 A

I_{2} + I_{4} = 4 + 2.78

I_{2} + I_{4} = 6.78 A

The sum of the I_{1} and I_{4} is 6.78 A. This current also flows through R_{8}.

V_{R8} = (I_{2} + I_{4})•R_{8}

V_{R8} = 6.78•8

V_{R8} = 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 I_{3}.

If V_{R3} = 10 V

I_{3} = | 10 | |

5 |

I_{3} = 2 A

Because of the voltage across R_{3} is 10 V, the voltage across R_{4} is 10 V too.

I_{4} = | 10 | |

10 |

I_{4} = 1 A

The current that flows through R_{5} is (I_{3} + I_{4})

V_{R5} = (I_{3} + I_{4})•R_{5}

V_{R5} = 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

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

3 |

R_{2,3,4} = 3 Ω

R_{1-5} = 6 + 3 + 1

R_{1-5} = 10 Ω

The equivalent resistance of the branching -2 is

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

R_{7,8,9} |

R_{7,8,9 }= | 12 | |

8 |

R_{7,8,9} = 1.5 Ω

R_{6,7,8,9} = 8.5 + 1.5 Ω

R_{6,7,8,9} = 10 Ω

The R_{1-5} and R_{6-9} are parallel.

R_{1-9} = | 10 | |

2 |

R_{1-9} = 5 Ω

The equivalent resistance of the circuit is

R_{s} = R_{1-9} + R_{10}

R_{s} = 5 + 7

R_{s} = 12 Ω

The current flowing from the voltage source is

I_{s} = | 48 | |

12 |

I_{s} = 4 A

Because of R_{1-5} and R_{6-9} are equal, the current flowing both arms are equal and 2 A.

I_{x} = | Is•R_{6-9} | |

R_{1-5} + R_{6-9} |

I_{x} = | 4•10 | |

20 |

I_{x} = 2 A

I_{y} = 4-2 = 2 A

The current flowing through resistor R_{3} is

I_{R3} = | 2•R_{3} | |

R_{2}+R_{3}+R_{4} |

I_{R3} = | 2*9 | |

27 |

I_{R3} = 0.667 A

V_{3} = 0.667•9

V_{3} = 6 V

Let we find the equivalent resistance of the resistors R_{7} and R_{9} for find the current that flows through R_{3}.

R_{7,9} = | 36 | |

15 |

R_{7-9} = 2.4 Ω

Now, we can find the current I_{8}.

I_{R8} = | 2•2.4 | |

6.4 |

I_{R8} = 0.75 A

V_{8} = I_{8}•R_{8}

V_{8} = 0.75•4

V_{8} = 3 V

V_{3} + V_{8 }= 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 R_{3} – R_{8} in the upper arm. The resistors R_{3} and R_{4} are parallel amongst themselves, the resistors R5 and R6 are parallel amongst themselves too.

R_{3,4} = | 6•12 | |

18 |

R_{3,4} = 4 Ω

R_{4,5} = | 8 | = 4 Ω |

2 |

R_{3-8} = 4 + 4 + 2 + 2

R_{3-8} = 12 Ω

The equivalent resistance of the branching -2 is

R_{1,2} = 3 + 5

R_{1,2} = 8 Ω

Branch -1 and branch-2 are parallel.

R_{s} = | R_{3-8} • R_{1-2} | |

R_{3-8} + R_{1-2} |

R_{s} = | 12•8 | |

20 |

R_{s} = 4.8 Ω

The main current is

I_{s} = | 96 | |

4.8 |

I_{s} = 20 A

The current flowing through branching -1 is

I_{x} = | I_{s}•R_{1,2} | |

R_{1,2} + R_{3-8} |

I_{x} = | 20•8 | |

20 |

I_{x} = 8 A

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

I_{3} = | I_{x}•R_{4} | |

R_{3} + R_{4} |

I_{3} = | 8•12 | |

18 |

I_{3} = 5.33 A

The right answer is option A

**Solution 5**

The equivalent resistance of the circuit is,

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

R_{s} = 6 + 3 + 5 + 1

R_{s} = 15 Ω

The current drawn from the supply source is

I_{s} = | 60 | = 4 A |

15 |

Suppose the resistor that we add to the circuit is R_{5} and R_{6}. The equivalent resistance of these resistors is

R_{5,6} = | 10 | |

2 |

R_{5,6} = 5 Ω

The resistor R_{5} will be connected in series to the circuit.

Now the equivalent resistance will be

R_{s} = 15 + 5

R_{s} = 20 Ω

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

I_{s} = | 60 | |

20 |

I_{s} = 3 A

The right answer is option E

**Solution 6**

We suppose that

R_{1} = 12 Ω

R_{2} = 15 Ω

R_{3} = 10 Ω

R_{4} = 3 Ω

R_{5} = 2 Ω

R_{1}, R_{2} and R_{3} are parallel. The resistors R_{3} and R_{4} will connect to these resistors.

The equivalent resistance of the R_{1}, R_{2} and R_{3}

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

R_{1,2,3} |

R_{1,2,3} = | 60 | |

15 |

R_{1,2,3} = 4 Ω

If it is connected two resistors R_{4} = 3 and R5 = 2 in series to the circuit, the equivalent resistance of the circuit is

R_{1-5} = 4 + 3 + 2

R_{1-5} = 9 Ω

The current drawn from the source is

I_{s} = | 36 | |

9 |

I_{s} = 4 A

The right answer is option B

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November 9 2018

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