An ammeter with a full scale deflection for \(I=10.0 \mathrm{A}\) has an internal resistance of \(24 \Omega .\) We need to use this ammeter to measure currents up to 12.0 A. The lab instructor advises that we get a resistor and use it to protect the ammeter. (a) What size resistor do we need and how should it be connected to the ammeter, in series or in parallel? (b) How do we interpret the ammeter readings?

To safely measure a current greater than the ammeter's full scale deflection, the resistor should be connected in parallel with the ammeter. This is because a parallel connection will split the current, allowing a portion of it to flow through the resistor, hence protecting the ammeter.

Since the ammeter has a full scale deflection of 10 A, we want to make sure that when it is connected in parallel with the resistor and a current of 12 A is applied, only 10 A flows through the ammeter. The remaining 2 A should flow through the resistor. We will use the current division rule to calculate the value of the resistor, R. The current division rule states the following: \(\frac{I_1}{I_2} = \frac{R_2}{R_1}\) In our case, \(I_1 = 10 A\) (ammeter current), \(I_2 = 2 A\) (resistor current), and \(R_1 = 24 Ω\) (ammeter resistance). Let R be the resistor value, which is \(R_2\). We can plug in the values and solve for R: \(\frac{10 A}{2 A} = \frac{R}{24 Ω}\)

Solving for R, we get: \(5 = \frac{R}{24 Ω}\) \(R = 5 \times 24 Ω\) \(R = 120 Ω\) So, the resistor needed is 120 Ω.

Since the 120 Ω resistor is connected in parallel with the ammeter, the current shown on the ammeter will be less than the actual current flowing through the circuit. To interpret the ammeter readings and obtain the actual current flowing through the circuit, we can use the following relationship: \(I_\text{actual} = \frac{R + R_\text{ammeter}}{R_\text{ammeter}} \times I_\text{ammeter}\) Where: \(I_\text{actual}\) = actual current in the circuit \(I_\text{ammeter}\) = current displayed on the ammeter \(R_\text{ammeter}\) = Ammeter's internal resistance (24 Ω) \(R\) = Resistor value (120 Ω) For example, if the ammeter reads 8 A, the actual current in the circuit will be: \(I_\text{actual} = \frac{120 Ω + 24 Ω}{24 Ω} \times 8 A = 6 \times 8 A = 48 A\) In summary, we need to connect a 120 Ω resistor in parallel with the ammeter to measure currents up to 12 A. To interpret the ammeter readings, we can use the above formula to find the actual current in the circuit.