# The Series Circuits Test-1 Solutions

## Calculations related to series connected electrical circuits. The equivalent resistance in series connected circuits. The equivalent current and voltage in series circuits. The solutions of the Test-1.

**Solution 1 **

First, let we find the current of the circuit. Because of the circuit is connected in series, it is flows the same current through all resistors.

To find the current, we must find the equivalent resistance of the circuit.

The equivalent resistance of the circuit.

R_{s} = R_{1} + R_{2} + R_{3} + R_{4} + R_{5} + R_{6}

R_{s} = 10 + 3 + 2 + 15 + 15 + 5

R_{s} = 50 Ω

I = V/R

I = 25/50

I = 0.5 A

V_{2} = I.R_{2}

V_{2} = 0.5 • 3

V_{2} = 1.5 V

V_{6} = I•R_{6}

V_{6} = 0.5 • 5

V_{6} = 2.5 V

V_{2} + V_{6} = 1,5 + 2,5 = 4 V

The right answer is option B.

**Solution 2**

The voltage that across any resistor in the series-connected circuits is founds as follow.

V_{R1} = | V_{s}•R_{1} | |

R_{s} |

V_{R1} : The voltage across R_{1}

V_{s}: Total voltage of the circuit.

R_{s}: The equivalent of the circuit.

Let we apply to the above question.

12 = | 50 • R_{2} | |

19 + R_{2} |

50R_{2} = 228 + 12R_{2}

38R_{2} = 228

R_{2} = 6 Ω

R_{s} = 19 + 6 = 25

I = 50/25 = 2 A

V_{2} = 6•2 = 12 V

The right answer is option C

**Solution 3**

The current of the circuit is found as follow.

I = | V | |

R_{s} |

I = 0.5 A ,

0.5 = | 110 | |

R_{s} |

R_{s} = | 110 | |

0.5 |

R_{s} = 220 Ω

The equivalent resistance of the circuit is 220 Ohm. Value of the resistor R_{4} is

R_{1} + R_{2} + R_{3} + R_{4} + R_{5} = 220

70 + 30 + 20 + R_{4} + 8 = 220

128 + R_{4} = 220

R_{4} = 92 Ω

The right answer is option A

**Solution 4 **

The equivalent resistance of the circuit is

R_{s} = 80 + 17 + 22 + 11 + 8 + 12

R_{s} = 150 Ω

Because the circuit is in series connected the same current flows through all resistor.

V_{3} = 8,8 V

V_{3} = I.R_{3}

_{}

8,8 = 22.I

I = 0,4 A

The current of the circuit is 0.4 A. The equivalent resistance is 150 Ω.

V_{s} = I_{s}.R_{s}

V = 0.4 • 150

V = 60 V

The right answer is option C

**Solution 5**

We can find the current of the circuit from value of V_{2} + V_{3} = 16 V

(R_{2} + R_{3}) • I = 16 V

40•I = 16 V

I = 0.4 A

R_{s} = 60 + 16 + 24 + 36 + 24

R_{s} = 160 Ω

V = 160 • 0.4

V = 64 V

The right answer is option D

**Solution 6**

The equivalent resistance of the circuit is

R_{s} = 30 + 17 + 18 + 11 + 34 + 27 + 23

R_{s} = 160 Ω

For I = 2 A

2 = | V | |

160 |

V = 320 V

Voltage supply should be 320 V.

The right answer is option C

**Solution 7**

I = | V | |

R |

V = 110 V

R = ?

R = 3 + 5 + R_{3} + 1 + 25 + 8 + 7

R = 49 + R_{3}

2 = | 110 | |

49 + R_{3} |

98 + 2R_{3} = 110

2R_{3} = 12

R_{3} = 6 Ω

The right answer is option E

**Solution 8**

Let we find the potential difference of the voltage source of the circuit.

V = I.R

V = 0.75 • 36

V = 27 V

If a resistor of 18 ohm is connected as series to the circuit, the equivalent resistance is

R_{s} = 36 + 18 = 54 Ω

The voltage is 27 V.

27 = 54•I

I = 0.5 A.

The right answer is option C

**Solution 9**

Let we find the voltage supply.

V = I•R

V = 2.5 • 60

V = 150 V

If we add the resistor R, current of the circuit will be 1.5 A.

1.5 = | 150 | |

60 + R |

90 + 1.5R = 150

1.5R = 60

R = 40 Ω

The right answer is option B

**Solution 10**

V = I•R

R = | V | |

I |

V = 24 V

I = 0.2 Ω

R = | 24 | |

0.2 |

R = 120 Ω

The right answer is option E

**Series Connected Circuits Subject Expression**

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21/10/2018

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