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1

You're just thinking 'upside down'. The 3 ohm resistor is just 1/4 the resistance of the 12 ohm resistor but, power is inversely proportional to the resistance. Thus, if you decrease the resistance by a factor of X, you increase the power by a factor of X. In this case, the resistance is decreased by a factor of 4 so the power is increased by a factor of ...


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what is your logic behind setting V=1? you do not have enough info to assume that. What you do know is that the voltage is constant (same cell), so you can write: $P_{ini}*R_{ini}=P_{final}*R_{final}$ Knowing both R's gets you $P_{final}=4P_{ini}$


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I know that if all the resistors were the same resistance than the way to get the max current to go through them and hence the max voltage and power would be to place them all in series, No, you place them in parallel. For a resistance, the power is given by $$p_R = \frac{V^2_R}{R}$$ Since the maximum voltage you have is 30V, the maximum power is ...


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The key think you need to know is that if you apply a voltage $V$ and a current $I$ flows then the power dissipated is: $$ P = VI \tag{1} $$ You know the power at 220V so you can calculate the current, and if we assume the kettle behaves as a simple resistor then it will obey Ohm's law so you can calculate the resistance. $$ R = \frac{V}{I} \tag{2} $$ ...


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First, the reactive power is not dissipated, but which corresponds to power delivered by the power stored in the reactive component (inductor or capacitor) during a semi-cycle; in the next half cycle, the component returns the stored energy to the source. For this to occur, the component must have the ability to store energy. In the case of a capacitor, the ...


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perhaps because a resistor (at least an ideal resistor) is not a reactive component. and neither do reactive components (such as capacitors and ideal inductors) dissipate power.


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Your logic is completely correct. Two 1/2 R resistors in parallel will dissipate 4x the power of a 1 R resistor at the same applied voltage. Whether that is "good" or not depends on whether the wire can handle the extra heat, whether the power supply can supply the extra current, whether the extra heat is desirable, whether using the extra power is ...


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yes, you will have to pay more if your load is inductive. most of energy meters work with the voltage and current to calculate energy. When we have inductive load it takes more current than resistive load to produce same power or output. P=V.I.Cosx where x is the phase angle between voltage(V) and current(I). if the phase angle x goes higher, the power ...


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A load is only purely inductive as long as you aren't taking any energy out of it. So a purely inductive load would certainly give us a low (zero!) electricity bill, but it wouldn't be much use as a domestic appliance since it couldn't do anything. Once your appliance starts doing any work it is no longer a purely inductive load. If an appliance does work, ...



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