Question

15.0 7.000 5.00 2.000

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?

Answer 1

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.