Maximizing Voltage and Current Ranges of Galvanometer Circuit Configurations

To determine which statements are correct, let us analyze the maximum voltage range and maximum current range for the given configurations.

Let:

• $I_g$ be the maximum full-scale deflection current for each identical galvanometer.
• $R_G$ be the internal resistance of each galvanometer (denoted as $R_C$ in the problem, where $R_G < \frac{R}{2}$).
• $R$ be the resistance of each identical resistor.

---

1. Maximum Voltage Range Analysis (Voltmeter Setup)

To maximize the voltage range, we need to maximize the potential difference across the terminal pair without exceeding the maximum full-scale deflection current $I_g$ through any galvanometer.

• Configuration A (All four components connected in series):

  • Total resistance of the circuit: $R_{\text{eq, A}} = 2R + 2R_G$
  • Since all components are in series, the maximum safe current through the circuit is $I_g$.
  • Maximum voltage range: $V_{\text{max, A}} = I_g R_{\text{eq, A}} = I_g (2R + 2R_G) = 2I_g (R + R_G)$

• Configuration B (Two galvanometers in parallel, connected in series with the two resistors):

  • Equivalent resistance of the parallel combination of two galvanometers: $R'_G = \frac{R_G}{2}$
  • Total equivalent resistance of the system: $R_{\text{eq, B}} = 2R + \frac{R_G}{2}$
  • The total current entering the system splits equally between the two identical parallel galvanometers. Therefore, to reach full-scale current $I_g$ in each galvanometer, the total current can be $2I_g$.
  • Maximum voltage range: $V_{\text{max, B}} = (2I_g) \cdot R_{\text{eq, B}} = 2I_g \left(2R + \frac{R_G}{2}\right) = 4I_g R + I_g R_G = I_g(4R + R_G)$

• Comparing $V_{\text{max, A}}$ and $V_{\text{max, B}}$: $V_{\text{max, B}} - V_{\text{max, A}} = I_g(4R + R_G) - I_g(2R + 2R_G) = I_g(2R - R_G)$ Given that $R_G < \frac{R}{2}$, it follows that $2R - R_G > \frac{3R}{2} > 0$.

  Therefore, $V_{\text{max, B}} > V_{\text{max, A}}$.

  Thus, Statement (B) is TRUE and Statement (A) is FALSE.

---

2. Maximum Current Range Analysis (Ammeter Setup)

To maximize the current range, we need to maximize the total current entering the setup such that no galvanometer experiences a current greater than $I_g$.

• Configuration C (All four components connected in parallel):

  • The maximum potential difference that can be applied across all parallel branches is limited by the maximum safe voltage across a single galvanometer: $V_{\text{max}} = I_g R_G$
  • At this maximum voltage:
    • Current through each galvanometer branch = $I_g$
    • Current through each resistor branch = $I_R = \frac{V_{\text{max}}}{R} = \frac{I_g R_G}{R}$
  • Total maximum current: $I_{\text{max, C}} = 2I_g + 2I_R = 2I_g + \frac{2I_g R_G}{R} = 2I_g \left(1 + \frac{R_G}{R}\right)$

• Configuration D (Two galvanometers in series, connected in parallel with both resistors):

  • Resistance of the series galvanometer branch = $2R_G$.
  • The maximum safe voltage across this galvanometer branch without exceeding current $I_g$ is: $V_{\text{max}} = I_g (2R_G) = 2I_g R_G$
  • At this voltage:
    • Current through the galvanometer branch = $I_g$
    • Current through each resistor branch = $I_R = \frac{V_{\text{max}}}{R} = \frac{2I_g R_G}{R}$
  • Total maximum current: $I_{\text{max, D}} = I_g + 2I_R = I_g + \frac{4I_g R_G}{R} = I_g \left(1 + \frac{4R_G}{R}\right)$

• Comparing $I_{\text{max, C}}$ and $I_{\text{max, D}}$: $I_{\text{max, C}} - I_{\text{max, D}} = 2I_g \left(1 + \frac{R_G}{R}\right) - I_g \left(1 + \frac{4R_G}{R}\right) = I_g \left(1 - \frac{2R_G}{R}\right)$ Given that $R_G < \frac{R}{2}$, we have $\frac{2R_G}{R} < 1$, which gives: $1 - \frac{2R_G}{R} > 0 \implies I_{\text{max, C}} > I_{\text{max, D}}$

  Therefore, the maximum current range is obtained when all components are connected in parallel.

  Thus, Statement (C) is TRUE and Statement (D) is FALSE.

---

Conclusion

The correct options are (B) and (C).