May i ask whether the current flow in a circuit depends upon the load or the voltage of the power source.For eg: In a circuit with a 12V battery and a lamp,does the current flow depend on the voltage of the battery or on the load(lamp) connected? |
by Tobinro
October 12, 2022 |

Will more current flow if the same lamp is connected to a 24V battery? |
by Tobinro
October 12, 2022 |

If there is a single resistive load, the current, I, is given by the voltage, V (ou U in some European countries) divided by the resistance, R, of the load: V = R I If you double V and keep R constant, the current, I, has to double too. For a lamp, the resistance changes with temperature. With tungsten, the resistance increases with the temperature, so, the current won't double because R will increases too because of the self-heating. |
by vanderghast
October 12, 2022 |

may i ask will the current flow in the circuit depend upon the power rating of the bulb.Like for eg: the current drawn by a 5w bulb is different from a 10w bulb right.According to the equation P=VI,for the same power rating as voltage increase,the current decreases right?could you advice on this please |
by Tobinro
October 13, 2022 |

The power rating is the maximum power that the component can dissipate safely without damaging it. A 1/8 Watt resistor of 100 ohm and a 1W resistor of 100 ohm will both draw a current of 35 mA from a battery of 3.5V. Sure, if you use, say, a pill cell battery, the battery may be unable to deliver that power and in that case, the circuit will be limited by the power that can deliver the battery. |
by vanderghast
October 13, 2022 |

Furthermore, while V = R I is an active constraint (an enforced constraint), P = V I <= Pmax is an inactive constraint for most circuits. It is a little bit like: x <= 3 and x^2 <= 100 And when P, of a component, reached or is greater than its rating, we can either change the component (e.g. take a 1W resistor instead of one at a rating of 1/2 W), or we redesign the circuit. |
by vanderghast
October 13, 2022 |

Could you let me know if power of an equipment has any role in determining the current through the circuit connecting it? |
by Tobinro
October 13, 2022 |

The power is not determining the current.
The power rating is the |
by vanderghast
October 13, 2022 |

May i ask if we are increasing the load in a circuit,i believe the current also increases.Does that increase in current caused by the additional power added to the circuit or due to the addition of the extra load in parallel which reduces the overall resistance which in turn increases the current? |
by Tobinro
October 17, 2022 |

"when we increase the load current increases because when load is increased power consumption increases and p=sq.root3.V.I.cos q. according to the expression, voltage & p.f. remains the same for a particular system. hence when load increase current increases"....May i ask if this statement is true? |
by Tobinro
October 17, 2022 |

Does resistance have any role in the increase in current in a circuit when the load is increased? |
by Tobinro
October 17, 2022 |

The current DECREASES when the load increases. V1 = V2, R2 > R1 and I2 < I1 |
by vanderghast
October 17, 2022 |

Thank you for the reply.Can i ask if i add more load(resistance) in parallel,will the current increase as the effective resistance is reduced?Is this current increase due to the effective decrease in the overall resistance or due to the increase in load power? |
by Tobinro
October 18, 2022 |

Adding load in parallel is like, under gravity, having more and more falling masses. You may add as many masses as possible that you want, they won't start falling faster because there are more of them. There will be more potential energy, in total, but each of the masses would fall independently. (Well, as long as their individual mass is "small" in comparison with Earth's mass.) So, for resistors, each resistor in parallel is independent of the other resistors in parallel. Sure THE SOURCE sees the sum of all the currents, as, for gravity, Earth will see all the masses. |
by vanderghast
October 18, 2022 |

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