each resistor in the circuit shown is 5.0 Ù, and the battery is 12.0 V, what does the second ammeter (A2) read?
Are there just two resistors and are they in parallel?
If so then the current through each leg is i = V/R = 12/5
To find the reading of the second ammeter (A2) in the circuit, we need to determine the total resistance of the circuit using Ohm's Law and then calculate the current flowing through the circuit.
Given that each resistor in the circuit is 5.0 Ω, and the battery voltage is 12.0 V, we can see that the circuit consists of three resistors connected in series.
Let's proceed step by step to get the answer:
Step 1: Calculate the total resistance (R_total) of the circuit.
Since the resistors are connected in series, we can simply add their resistances.
R_total = R1 + R2 + R3 = 5.0 Ω + 5.0 Ω + 5.0 Ω = 15.0 Ω
Step 2: Determine the current (I) flowing through the circuit using Ohm's Law.
Ohm's Law states that the current (I) is equal to the voltage (V) divided by the resistance (R): I = V / R.
In this case, the battery voltage is 12.0 V and the total resistance is 15.0 Ω.
I = 12.0 V / 15.0 Ω ≈ 0.8 A
Step 3: Find the current flowing through the second ammeter (A2).
Since the second ammeter (A2) is connected in series with the battery and the first resistor, the current passing through it will be the same as the total circuit current.
Therefore, the second ammeter (A2) will read approximately 0.8 A.
So, the reading of the second ammeter (A2) in the circuit is approximately 0.8 A.