Question

An ammeter with an internal resistance of \(53.0 \Omega\) measures a current of \(5.25 \mathrm{~mA}\) in a circuit containing a battery and a total resistance of \(1130 \Omega\). The insertion of the ammeter alters the resistance of the circuit, and thus the measurement does not give the actual value of the current in the circuit without the ammeter. Determine the actual value of the current.

Short answer

Answer: The actual value of the current in the circuit without the ammeter is 5.49 mA.

Step-by-step solution

Calculate total voltage without ammeter

We know that the ammeter measures a current of 5.25 mA, and its internal resistance is 53.0 Ω. Using Ohm's Law, we can calculate the voltage across the ammeter (V_am): V_am = I_am * R_am Where I_am is the current through the ammeter and R_am is the ammeter's internal resistance. V_am = 5.25 * 10^(-3) A * 53.0 Ω = 0.27825 V

Calculate total voltage in the circuit

Now, we need to calculate the total voltage in the circuit (V_total), considering that the ammeter has increased the total resistance of the circuit to (1130 Ω + 53.0 Ω) = 1183 Ω. We will use Ohm's Law again: V_total = I_read * R_total Where I_read is the current reading from the ammeter (5.25 mA), and R_total is the total resistance of the circuit with the ammeter inserted (1183 Ω). V_total = 5.25 * 10^(-3) A * 1183 Ω = 6.21075 V

Calculate the actual current in the circuit

Finally, we can find the actual current in the circuit (I_actual) without the ammeter by using Ohm's Law, along with the original circuit resistance (1130 Ω) and the total voltage we calculated in Step 2 (V_total): I_actual = V_total / R_original Where R_original is the original resistance of the circuit without the ammeter. I_actual = 6.21075 V / 1130 Ω = 5.49 mA The actual value of the current in the circuit without the ammeter is 5.49 mA.