# Alternating Current R Circuit

## Alternating current circuits. The circuits R that occurs only one resistor. Finding of the maximum current and maximum voltage in the alternating currents. The subject expression and examples.

An alternative current circuit consists a sinusoidal voltage source, resistors, bobines, condansators. In an AC circuit can found one or a few of bobbins, resistors and condensators.

The voltage source that supply the circuit is a sinusoidal voltage source. So, this source produces alternative current. Alternative current is a current that switched its polarity over time. The elements of an alternating current circuit are resistor, bobbin and condansator. An alternating current circuit can be occurs from only one source and one resistor or can consists all of them (resistor, bobbin, condansator).

In this lesson we will do the calculates related to circuits that occurred only one source an done resistor.

The voltage at any t time in the AC circuit is found by the following formula.

V = V_{max }• sinωt

The alternating voltage is voltage that its magnitude switching over time. The magnitude of the AC voltage is be 0 V twice in one period, and it is reach its maximum value in twice of one period.

**The R Circuits**

The R circuits occurs from only a voltage source and resistor. There are not condensator and bobbins in these circuits.

In the above figure is shown an alternating current circuit that occurs from only one resistor and one source.

The value at any time of the voltage across resistor R is given by the following formula.

V = Vm • sinωt

The value at any time of the current flowing through the resistor R is given by the following formula.

I = | V_{max}•sinωt | |

R |

The graph of the voltage across the resistor R and the current flowing through has given in the following figure. In this circuit the current and the voltage are in the same phases.

**Examples:**

The circuit in the above figure has operates in frequency of 50 Hz.

a) Find the maximum voltage and the maximum current for this circuit.

b) Find the value of the current and voltage in the t = 5 ms.

(sin108° = 0.95)

**Solution:**

The maximum value of the voltage is the voltage at time that value of the sinωt is 1. At this time the maximum value of the voltage is be 300 V .

Maximum value of the current is

I_{max} = | V_{max} | |

R |

I_{max} = | 300 | |

50 |

I_{max} = 6 A olur.

b)

V = 300•sin(2•π•f•t)

Let we find tha value of sin(2•π•f•t)

Sin(2•π•60•0.005) = sin(0.6•π)

Let we find the value of 0.6π. The π is equal to 180° .

0.6π = 0.6•180 = 108°

The value of the voltage is 300.sin108°.

V = 300•sin108°

V = 300•0.95

V = 285

I = | 285 | |

50 |

I = 5.7 A

**Example:**

Find the effective value of the current and voltage for the circuit shown the figure above .

**Solution:**

The maximum value of the voltage in the circuit is 240 V. Effective value of this voltage is

V_{eff} = | V_{max} | |

√2 |

V_{eff} = | 240 | |

0.707 |

V_{eff} = 339.5 V ≅ 340 V

Effective value of the current is

I_{eff}= | V_{eff } | |

R |

I_{eff }= | 340 | |

170 |

I_{eff} = 2 A

**Example:**

An alternating voltage is given with 150sin(100πt).

A - How much Hertz is the frequency of this voltage.

B- How much Volts is the effective value of this voltage.

Solution:

The angular velocity is

ω= 2•π•f

2•π•f = 100π

f = 50 Hz

Effective value of the voltage is

V_{eff} = | 150 | |

√2 |

V_{eff} ≅ 106 V

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January 22 2019

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