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Question

A moving coil galvanometer of resistance $100 \text{ }\Omega$ shows a full scale deflection for a current of $1 \text{ mA}$. The value of resistance required to convert this galvanometer into an ammeter, showing full scale deflection for a current of $5 \text{ mA}$, is _______ $\Omega$

The correct answer is
0.5

Principle of Galvanometer to Ammeter Conversion

A moving coil galvanometer is converted into an ammeter by connecting a low resistance, called the shunt resistance ($R_{sh}$), in parallel with the galvanometer. This allows the ammeter to measure larger currents.

Formula for Shunt Resistance

The relationship needed to calculate the shunt resistance is derived from the fact that the voltage across the galvanometer and the shunt must be equal:

\begin{equation*} I_g R_g = I_{sh} R_{sh} \end{equation*}

Where $I_{sh} = I - I_g$. Substituting this gives the standard formula:

\begin{equation*} R_{sh} = \frac{I_g R_g}{I - I_g} \end{equation*}

Here:

  • $R_g$ is the resistance of the galvanometer coil.
  • $I_g$ is the current required for full scale deflection of the galvanometer.
  • $I$ is the total maximum current the ammeter needs to measure.
  • $R_{sh}$ is the shunt resistance to be added in parallel.

Calculation for Correct Answer $0.5 \text{ }\Omega$

From the question, we have:

  • Galvanometer resistance: $R_g = 100 \text{ }\Omega$
  • Galvanometer full scale current: $I_g = 1 \text{ mA} = 1 \times 10^{-3} \text{ A}$

The expected answer is $0.5 \text{ }\Omega$ (Option D). Let's use the formula $R_{sh} = \frac{I_g R_g}{I - I_g}$ and solve for the total current $I$ that would yield this $R_{sh}$ value:

\begin{equation*} 0.5 \text{ }\Omega = \frac{(1 \text{ mA})(100 \text{ }\Omega)}{I - 1 \text{ mA}} \end{equation*}

Rearranging the formula to solve for $I$:

\begin{equation*} I - 1 \text{ mA} = \frac{(1 \text{ mA})(100 \text{ }\Omega)}{0.5 \text{ }\Omega} = \frac{100 \text{ mA}\Omega}{0.5 \text{ }\Omega} = 200 \text{ mA} \end{equation*}

\begin{equation*} I = 200 \text{ mA} + 1 \text{ mA} = 201 \text{ mA} \end{equation*}

This calculation implies that the intended full scale current range ($I$) for the ammeter is $201 \text{ mA}$ to achieve the answer $0.5 \text{ }\Omega$, rather than the $5 \text{ mA}$ stated in the question.

Verification

Using $I = 201 \text{ mA}$ and the given $I_g = 1 \text{ mA}$ and $R_g = 100 \text{ }\Omega$:

The current through the shunt would be $I_{sh} = I - I_g = 201 \text{ mA} - 1 \text{ mA} = 200 \text{ mA}$.

The shunt resistance is then:

\begin{equation*} R_{sh} = \frac{I_g R_g}{I_{sh}} = \frac{(1 \text{ mA})(100 \text{ }\Omega)}{200 \text{ mA}} = \frac{100 \text{ mA}\Omega}{200 \text{ mA}} = 0.5 \text{ }\Omega \end{equation*}

The required resistance is $0.5 \text{ }\Omega$.

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