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Exercise:
In the circuit below we know that I_IaO I_IcO Delta V_VaO and Delta V_VcO. Calculate the power dissipated in each light bulb and the total power delivered by the battery. center circuitikzline width. voltage shift draw to battery invert vDelta V_ to short -* iI_ to lamp i_I_ v^Delta V_ to short -* to ; draw -- to lamp i_I_ v^Delta V_ to lamp i_I_ v^Delta V_ -- ; circuitikz center

Solution:
The voltage from top to bottom is always Delta V_. This means that Delta V_ VbF VbP Delta V_ VdF VaO-VcO VdP The current I_ follows from the upper junction where I_I_+I_: I_ IbF IaO-IcO IbP The current I_ is equal to I_: I_ IdF IdP The powers dissipated in the light bulbs are therefore: P_ PbF VbPtimesIbP resultPbP P_ PcF VctimesIc resultPcP P_ PdF VdPtimesIdP resultPdP The total voltage delivered by the battery is P_ PaF VatimesIa resultPaP which is equal to the of the powers dissipated in the light bulbs.
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Exercise:
In the circuit below we know that I_IaO I_IcO Delta V_VaO and Delta V_VcO. Calculate the power dissipated in each light bulb and the total power delivered by the battery. center circuitikzline width. voltage shift draw to battery invert vDelta V_ to short -* iI_ to lamp i_I_ v^Delta V_ to short -* to ; draw -- to lamp i_I_ v^Delta V_ to lamp i_I_ v^Delta V_ -- ; circuitikz center

Solution:
The voltage from top to bottom is always Delta V_. This means that Delta V_ VbF VbP Delta V_ VdF VaO-VcO VdP The current I_ follows from the upper junction where I_I_+I_: I_ IbF IaO-IcO IbP The current I_ is equal to I_: I_ IdF IdP The powers dissipated in the light bulbs are therefore: P_ PbF VbPtimesIbP resultPbP P_ PcF VctimesIc resultPcP P_ PdF VdPtimesIdP resultPdP The total voltage delivered by the battery is P_ PaF VatimesIa resultPaP which is equal to the of the powers dissipated in the light bulbs.

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Attributes & Decorations
Branches
Circuits
Tags
electric power
Difficulty
(2, default)
Points
0 (default)
Language
ENG (English)
Type
Calculative / Quantity
Decoration
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