Chapter 24, Problem 75A

a.

To determine

The potential difference across the galvanometer for a given current flow through it.

Answer to Problem 75A

The potential difference across the galvanometer is $\text{50 mV}$.

Explanation of Solution

Given:

A galvanometer with a resistor connected in parallel to it. Its full-scale deflection corresponds to a current of 10-mA. The current through the galvanometer is $I_g=50\mu \text{A}$ and its resistance is $R_g=1\text{ k}\Omega$ .

Formula used:

Consider the circuit shown in Figure 1. The circuit consists of a galvanometer which can be used to measure small currents. The resistor connected in parallel with the galvanometer is called as “shunt resistor”. This resistor has a value less than the resistance of the galvanometer. It is converted to an ammeter using circuit.

  

In this circuit, the voltage across both the galvanometer and the shunt resistor are the same. The current through the galvanometer is given by,

  $I_g=\frac{V_s}{R_g}$

From this the potential difference across the galvanometer can be calculated as,

  $V_s=I_g{R}_g$ ……. (1)

Calculation:

Substituting the current through the galvanometer $I_g=50\mu \text{A}$ and the its resistance $R_g=1\text{ k}\Omega$ , the potential difference across the galvanometer is obtained using (1) as,

  $\begin{array}{c}{V}_s=50\times {10}^{-6}\times 1\times {10}^3\\ =50\text{ mV}\end{array}$

Conclusion:

The potential difference across the galvanometer is 50 mV.

b.

To determine

The equivalent resistance value

Answer to Problem 75A

The equivalent resistance of the circuit is $\text{5 }\Omega$

Explanation of Solution

Given:

A galvanometer with a resistor connected in parallel to it. The total current in the circuit is 10 mA.

Formula used:

Consider the circuit shown in Figure 1. In this circuit, since the voltage across both the galvanometer and the shunt resistor are the same, the value of total resistance can be calculated as,

  $R=\frac{V_s}{I}$

The total current in the circuit I is the sum of current through the galvanometer $I_g$ and that through the shunt resistor $I_{sh}$ .

Calculation:

Substituting the potential difference obtained in the previous part of $V_s=50\text{ mV}$ and the total current through the circuit $I=10\text{ mA}$ , the equivalent resistance of the circuit in Figure 1 is obtained as,

  $\begin{array}{c}R=\frac{50\times {10}^{-3}}{10\times {10}^{-3}}\\ \text{=5 }\Omega \end{array}$

Conclusion:

The equivalent resistance of the circuit is $\text{5 }\Omega$ .

c.

To determine

The shunt resistor value.

Answer to Problem 75A

The resistance of the shunt resistor is $\simeq \text{5 }\Omega$.

Explanation of Solution

Given:

A galvanometer with a resistor connected in parallel to it. The total current in the circuit is 10 mA.

Formula used:

Consider the circuit shown in Figure 1. In this circuit, since the voltage across both the galvanometer and the shunt resistor are the same, the value of shunt resistor can be calculated as,

  $R_{sh}=\frac{V_s}{I_{sh}}$

The total current in the circuit I is the sum of current through the galvanometer $I_g$ and that through the shunt resistor $I_{sh}$ The current through the shunt resistor can be found provided the other two currents are given as,

  $I_{sh}=I-I_g$

Calculation:

Substituting the total current through the circuit and the current through the galvanometer, the shunt current can be calculated as,

  $\begin{array}{c}{I}_{sh}=10\times {10}^{-3}-50\times {10}^{-6}\\ =9.95\text{ mA}\end{array}$

Substituting the potential difference obtained in the previous part of $V_s=50\text{ mV}$ and the current through the shunt resistor $I_{sh}=9.95\text{ mA}$ , the shunt resistance of the circuit in Figure 1 is obtained as,

  $\begin{array}{c}{R}_{sh}=\frac{50\times {10}^{-3}}{9.95\times {10}^{-3}}\\ \text{=5}\text{.025 }\Omega \\ \simeq 5\Omega \end{array}$

Conclusion:

The resistance of the shunt resistor is $\text{5 }\Omega$ .