In the circuit shown in the figure, R₁ = R₂ = 90.0 Ω, R₃ = R₄ = 20.0 Ω, V₁ = 7.0 V, V₂ = 8.0 V, and the batteries are both ideal. What current does the ammeter read? A) 0.40 A B) 0.050 A C) 0.83 A D) 0.056 A

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Using Kirchhoff's laws, we can solve for the current in the circuit. First, we can find the total resistance in the circuit by adding up the resistances in series and parallel. R₁ and R₂ are in series, so their total resistance is: RSeries = R₁ + R₂ = 90.0 Ω + 90.0 Ω = 180.0 Ω R₃ and R₄ are in parallel, so their total resistance is: RParallel = (R₃ x R₄) / (R₃ + R₄) = (20.0 Ω x 20.0 Ω) / (20.0 Ω + 20.0 Ω) = 10.0 Ω Now we can find the total resistance in the circuit: RTotal = RSeries + RParallel = 180.0 Ω + 10.0 Ω = 190.0 Ω Using Ohm's law, we can find the total current in the circuit: ITotal = (V₁ + V₂) / RTotal = (7.0 V + 8.0 V) / 190.0 Ω = 0.083 A Finally, we can use Kirchhoff's current law at the bottom node to find the current through the ammeter: IAmmeter = ITotal - (V₃ / R₃) - (V₄ / R₄) IAmmeter = 0.083 A - (20.0 V / 20.0 Ω) - (20.0 V / 20.0 Ω) = 0.4 A Therefore, the answer is A.

In the Circuit Shown in the Figure, R1 = R2

In the Circuit Shown in the Figure, R₁ = R₂