In the circuit shown in (Figure 1) , current flows through the 5.00 Ω resistor in the direction shown, and this resistor is measured to be consuming energy at a rate of 24.9 W . The batteries have negligibly small internal resistance.
Question: What current does the ammeter A read?
let current through the 5 ohm resistor is i ampere
energy consume rate = i^2*R = 24.9 W
=> i^2*5 = 24.9
=> i = 2.231 A
assuming right end of the 5 ohm resistor is at zero potential
=> left end potential = 0 + i*R = 2.231*5 = 11.15 V
SO current through the Ammeter branch = (15 - 11.15)/7 = 0.55 A leaving the 15 V battery.
In the circuit shown in (Figure 1) , current flows through the 5.00 Ω resistor in...
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