Chapter 19, Problem 59P

To determine

The actual current in the absence of ammeter.

Answer to Problem 59P

The actual current in the absence of ammeter is $5.52\times {10}^{-3}\text{A}$ .

Explanation of Solution

Given info:

The internal resistance of the ammeter is, $r=63\Omega$ .

The resistance of the first resistor is, $R_1=750\Omega$ .

The resistance of the second resistor is, $R_2=480$ .

The reading of the ammeter is, $I_a=5.25\text{mA}=5.25\times {10}^{-3}\text{A}$ .

Formula Used:

The expression to calculate the actual current in the absence of ammeter is,

  $I_c=\frac{I\left(R_1+R_2+r\right)}{R_1+R_2}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{I}_c=\frac{\left(5.25\times {10}^{-3}\text{A}\right)\left(750\Omega +480\Omega +63\Omega \right)}{750\Omega +480\Omega }\\ =5.52\times {10}^{-3}\text{A}\end{array}$

Conclusion:

Thus, the actual current in the absence of ammeter is $5.52\times {10}^{-3}\text{A}$ .