Chapter 24, Problem 95A

a.

To determine

The shunt resistor for the $50\mu \text{A}$ meter.

Answer to Problem 95A

Shunt resistor, $\text{R=0}\text{.428 }\Omega$ .

Explanation of Solution

Given:

Full scale Current, $I=50\mu \text{A}$

Resistance, $R=855\Omega$

Formula Used:

  $R=\frac{V}{I}$

V is the voltage

R is the resistance

Iis the current.

Calculations:

The voltage across the meter coil at full scale

  $\begin{array}{c}V=IR\\ =\left(50\mu \text{A}\right)\left(855\Omega \right)\\ =0.0428\text{}\text{ V}\end{array}$

Shunt resistance, $R=\frac{V}{I}$

  $R=\frac{0.0428\text{}\text{ V}}{100\text{ mA-}50\mu \text{A}}$

  $R=0.\text{428 }\Omega$

Conclusion:

Shunt resistor, $R=0.\text{428 }\Omega$ .

b.

To determine

The shunt resistor for the $500\mu \text{A meter}$ .

Answer to Problem 95A

Shunt resistor, $R=\text{4}\text{.30 }\Omega$ .

Explanation of Solution

Given:

Full scale Current, $I=500\mu \text{A}$

Resistance, $R=855\Omega$

Formula Used:

  $R=\frac{V}{I}$

V is the voltage

R is the resistance

Iis the current.

Calculations:

The voltage across the meter coil at full scale

  $\begin{array}{c}V=IR\\ =\left(500\mu \text{A}\right)\left(855\Omega \right)\\ =0.428\text{}\text{ V}\end{array}$

Shunt resistance, $R=\frac{V}{I}$

  $R=\frac{0.428\text{}\text{ V}}{100\text{ mA-}50\mu \text{A}}$

  $R=\text{4}\text{.30 }\Omega$

Conclusion:

Shunt resistor, $R=\text{4}\text{.30 }\Omega$ .

c.

To determine

To Explain: The galvanometer which is better for using.

Explanation of Solution

Introduction:

A galvanometer is an instrument used to measure the current in the circuit. It is used in series to measure the current and ideally has nil resistance.

The $50\mu \text{A}$ meter is better for use as it has lower resistance. The lower resistance will do less to alter the current of the circuit and measure a more accurate value of the current.

Conclusion:

The $50\mu \text{A}$ meter is better for use as it has lower resistance.