Question

The ammeter your physics instructor uses for in-class demonstrations has internal resistance \(R_{\mathrm{i}}=75.0 \Omega\) and measures a maximum current of \(1.50 \mathrm{~mA}\). The same ammeter can be used to measure currents of much greater magnitudes by wiring a shunt resistor of relatively small resistance, \(R_{\text {shunt }}\) in parallel with the ammeter. (a) Sketch the circuit diagram, and explain why the shunt resistor connected in parallel with the ammeter allows it to measure larger currents. (b) Calculate the resistance the shunt resistor has to have to allow the ammeter to measure a maximum current of 15.0 A

Short answer

Answer: The required resistance of the shunt resistor is approximately 0.0075 Ohms.

Step-by-step solution

Sketch the circuit diagram and explain the role of the shunt resistor

The shunt resistor should be connected in parallel with the ammeter, as shown in the circuit diagram below: ─────────┬────┬────────── │ │ │ ├─────╗ R_shunt │ │ R_i ┤ │ (ammeter)│ │ │ │ │ ├─────╝ R_shunt │ │ ├────┴────────── By connecting the shunt resistor in parallel with the ammeter, it allows the current to be split between the ammeter (with internal resistance R_i) and the shunt resistor. This reduces the current flowing through the ammeter, thus preventing damage to the device and allowing it to measure larger currents.

Calculate the resistance of the shunt resistor

Let's denote the total current as I_total = 15.0 A and the maximum current measured by the ammeter alone as I_ammeter = 1.50 mA. We are trying to find the value of R_shunt, which will allow the ammeter to measure the maximum current of 15.0 A. Using Kirchhoff's current law, we can say: I_total = I_ammeter + I_shunt We know that I_ammeter and I_total. We can calculate I_shunt using this information: I_shunt = I_total - I_ammeter = 15.0 A - 1.50 mA = 15.0 A - 0.0015 A = 14.9985 A Now, we can use Ohm's law to find the shunt resistor's resistance: V = I * R Since the ammeter and shunt resistor are in parallel, the voltage across both will be the same. Therefore: V_ammeter = V_shunt I_ammeter * R_i = I_shunt * R_shunt We know R_i = 75.0 Ohms and we have calculated I_shunt and I_ammeter earlier. Thus, we can solve for R_shunt: R_shunt = (I_ammeter * R_i) / I_shunt = (0.0015 A * 75.0 Ohms) / 14.9985 A ≈ 0.0075 Ohms So, the shunt resistor should have a resistance of approximately 0.0075 Ohms to allow the ammeter to measure a maximum current of 15.0 A.