© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Known Quantities
• Full-scale current of the galvanometer, $I_g = 10^{-4}\,\text{A}$.
• Series resistance to form a 0−5 V voltmeter, $R_S = 2 \times 10^6\,\Omega$ (2 MΩ).
• Desired voltmeter range, $V_{\max} = 5\,\text{V}$.
• Desired ammeter range, $I_{\max} = 0.10\,\text{mA} = 0.10 \times 10^{-3}\,\text{A} = 10^{-4}\,\text{A}$.
• Let the internal resistance of the galvanometer be $R_G$.
• We wish to find the shunt resistance $R_{\text{sh}}$ needed to convert the galvanometer into an ammeter of range $0.10$ mA.
Step 2: Use Voltmeter Condition to Determine $R_G$
When the galvanometer is converted into a voltmeter with range 5 V, the current through the galvanometer at full scale (which is $10^{-4}\,\text{A}$) flows through both $R_S$ and $R_G$. Hence, the total voltage across them must be:
$$
I_g \times (R_S + R_G) = V_{\max}.
$$
Substituting the known values:
$$
10^{-4} \times (2 \times 10^6 + R_G) = 5.
$$
Simplify:
$$
10^{-4} \times 2 \times 10^6 + 10^{-4} \times R_G = 5
$$
$$
\Rightarrow 200 + 10^{-4} R_G = 5
$$
$$
\Rightarrow 10^{-4} R_G = 5 - 200 = -195.
$$
Therefore,
$$
R_G = -195 \times 10^4\,\Omega,
$$
which is clearly a negative value for resistance.
Step 3: Interpret the Result
A negative internal resistance $R_G$ is physically impossible. Thus, the given data about converting the same galvanometer to a 5 V range must be inconsistent or the question’s conditions are self-contradictory.
Step 4: Conclusion Regarding the Ammeter Shunt
Because $R_G$ cannot be negative, none of the listed options for the shunt resistance (200 Ω, 500 Ω, 100 Ω) can be correct under these conditions. Hence, the correct answer must be that none of the provided options is valid.
Answer: None of the options are correct.
