# Solutions Of The Electric Power Test II

## Subject of electrical circuits. The consumed power in serial and parallel circuits. Total power taken from voltage source. Solutions of the solved questions II.

**Solution 1**

If we know the power in R_{6}, we can find the current I_{6} using this power.

If we can find the current I_{6}, we can find the current I_{3} too.

P_{6} = 16 W

P_{6} = (I_{6})^{2}*R_{6}

16 = (I_{6})^{2}*4

(I_{6})^{2} = 4

I_{6} = 2 A

I_{6} = | I_{3}*12 | |

12+6 |

2 = | I_{3}*12 | |

18 |

36 = 12*I_{3}

I_{3} = 3 A

The right answer is option B

**Solution 2**

We can find the current flow R_{3} and the voltage drop across R_{3} using the power in the R_{3}.

If we can find the current Ix that flows in this arm, we can find the current Iy.

If we can find the equivalent resistance of circuit and total current of the circuit, we can calculate the voltage of the supply source.

P_{3 }= 24 W

24 = (I_{3})^{2}*R_{6}

24 = (I_{3})^{2}*6

I_{3} = 2 A

V_{R3} = 2*6

V_{R3} = 12 V

The voltage drops in parallel arms are equal.

If resistance and voltage in the arms are known, the current flowing through these arms can be find.

V_{R4} = V_{R2} = V_{R3 }

V_{R4} = 12 V

I_{4} = | 12 | |

12 |

I_{4} = 1 A

V_{R2} = 12 V

I_{2} = | 12 | |

4 |

I_{2} = 3 A

Total current of lower arm is

I_{x} = I_{2} + I_{3} + I_{4}

I_{x} = 3 + 2 + 1

I_{x} = 6 A

The voltage drop across R_{1} is

V_{R1} = I_{x}*R_{1}

V_{R1} = 6*6

V_{R1} = 36 V

The voltage drop in the lower arm is equal to the sum of the V_{R1} with voltage that one of the parallel arms.

V_{bottom} = V_{R1} + V_{R4}

V_{bottom} = 12 + 36 = 48 V

Because of the voltage drops across parallel arms are equal, the voltage of the upper arm is 48 V too.

Let’s find the equivalent resistance of the upper arm.

R_{6,7} = | 20*5 | |

25 |

R_{6,7} = 4 Ω

R_{5-7} = 4 + 4 = 8 Ω

I_{y} = | 48 | |

8 |

I_{y} = 6 A

The current I_{8} is equal the sum of the currents I_{x} and I_{y}.

I_{8} = I_{x} + I_{y} = 6 + 6

I_{8} = 12 A

V_{R8} = 12*6 = 72 V

Total voltage = V_{lower} + V_{R8}

V_{source} = 48 + 72

V_{source} = 120 V

The right answer is option E

**Solution 3**

The equivalent resistance of the Branching -2 is

R_{7,8} = | 6*4 | |

10 |

R_{7,8} = 2,4 Ω

R_{B2} = R_{6} + R_{7,8} + R_{9}

R_{B2} = 2 + 2,4 + 1,6

R_{B2} = 6 Ω

The equivalent resistance of the Branching -1 is

R_{4,5} = | 12*8 | |

20 |

R_{4,5} = 4,8 Ω

R_{B1} = R_{3} + R_{4,5}

R_{B1} = 7,2 + 4,8 = 12 Ω

Branch -1 and Branch-2 are parallel. The equivalent resistance of these branches is,

R_{B1-2} = | 12*6 | |

18 |

R_{B1-2} = 4 Ω

Total resistance of the circuit is

R_{s} = R_{1} + R_{2} + R_{B1-2}

R_{s } = 3 + 5 + 4

R_{s }= 12 Ω

Main current of the circuit is

I_{s} = | 48 | |

12 |

I_{s} = 4 A

I_{x} = I_{s} = 4 A

I_{z} = | I_{x}*R_{B1} | |

R_{B1} + R_{B2} |

I_{z} = | 4*12 | |

18 |

I_{z} = 2,67 A

I_{8} = | I_{z}*R_{7} | |

R_{7}+R_{8} |

I_{8} = | 2,67*6 | |

10 |

I_{8} = 1,6 A

P_{8} = (I_{8})^{2}*R_{8}

P_{8} = 2,56*4

P_{8} = 10,24 W

The right answer is option C

**Solution 4**

First we must find the equivalent resistance of the circuit. Then we find main current of the circuit. We can find the power drawn from the source by equality of the V*I or I^{2}R.

R_{3-5} = 30 Ω

R_{1-2} = 30

R_{1-5} = | 30 | = 15 Ω |

2 |

R_{1-8} = 15 + 2 + 5 + 8

R_{1-8} = 30 Ω

R_{s} = 30 Ω

I_{s} = | 60 | |

30 |

I_{s} = 2 A

P = V*I

P = 60*2

P = 120 W

The right answer is option A

**Solution 5**

If three resistors of 16 ohms are connected in parallel to a 48 V source, the voltage drop at each resistor is 48 V.

Current flowing through each resistor is

I_{1} = I_{2} = I_{3}

I_{1} = | 48 | |

16 |

I_{1} = 3 A

I_{2} = 3 A

I_{3} = 3 A

Total current drawn from source is

I_{1} + I_{2} + I_{3} = 9 A

Power in each resistors is

P_{1} = (I_{1})^{2}*R_{1}

P_{1} = 9*16 = 144 W

If three resistors are parallel and equal to each other, total power drawn from source is found by multipliying of the power of one of resistor by 3.

P_{t} = 144*3 = 432 W

If an 8 Ohm resistor is connected to these resistors in parallel, the voltage drop across this resistor is be 48 V too. Suppose the current flowing through this resistor is I4.

I_{4} = | 48 | |

8 |

I_{4} = 6 A

In this case, there is no any change in the current and power of the 3 resistor of 16 Ohm . Only, due to fourth resistance an additional current and power is transmitted to the circuit.

Power in the resistor of 8 ohm is

P_{4} = 36*8 = 288 W

The total power will increase 288 W.

The right answer is option B

**Solution 6**

The equivalent resistance of three resistors of 8,12 and 15 Ohm is

R_{s} = 18 + 12 + 15

R_{s} = 45 Ω

In this case, current drawn from the source is

I_{s} = | 90 | |

45 |

I_{s} = 2 A

In this case, total power drawn from circuit is

P_{t} = V*I

P_{t} = 90*2 = 180 W

In addition, if a resistor of 15 Ohm is connected in series to these resistors, the equivalent resistance of the circuit is

R_{s} = 45 + 15

R_{s} = 60 Ω

In this case, the total current is

I_{s} = | 90 | |

60 |

I_{s} = 1,5 A

In last case, total power drawn from source is

P_{t} = V*I

P_{t} = 90*1,5

P_{t} = 135 W

Difference = 180 – 135

Difference = 45 W

The power drawn from source has been decreased 45 W.

The right answer is option C

**Power Questions In Electric Circuits Test-1**

RISE KNOWLEDGE

November 28 2018

- WRITE COMMENT
- NAME SURNAME(or nick)
- COMMENT