Calculate Current Drawing at 120V After 15.8% Voltage Reduction

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To calculate the current drawn at 120V after a 15.8% voltage reduction, the resistance of the device must be adjusted to reflect the reduced voltage. The original current is 2.56 A at 120 V, with a resistance of 46.875 ohms. A voltage drop of 15.8% means the new voltage is 101.4 V (85% of 120 V). The correct approach involves using Ohm's Law, where the new current is calculated as 101.4 V divided by the adjusted resistance, leading to a current of approximately 2.16 A. Understanding the percentage reduction in voltage and its impact on current is crucial for accurate calculations.
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Homework Statement


If the resistance of the device were reduced by 15.8 percent, what current would be drawn at 120 V?

there was a first part of this question which i have already gotten. the question was
An electrical device draws 2.56 A at 120 V. If the voltage drops by 15.8 percent, what will be the current, assuming nothing else changes? so the resistance is 46.875


Homework Equations


ok so i believe that i would just do 120/(.158x46.875)


The Attempt at a Solution


when i did this i got like 16 A and it was said to be wrong. i am confused on what i did wrong.
 
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Reducing by 15% means you keep 85% of the resistance. Multiply by .842, not .158.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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