© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Key Quantities
• The galvanometer has 150 divisions in total.
• Current sensitivity = 10 divisions per mA (i.e., for 1 mA, it deflects 10 divisions).
• Voltage sensitivity = 2 divisions per mV (i.e., for 1 mV, it deflects 2 divisions).
• We want each division to read 1 volt, so for full-scale (150 divisions) the galvanometer must correspond to 150 volts.
Step 2: Compute the Galvanometer’s Internal Resistance
For 1 division, the current needed is 0.1 mA (because 10 divisions per mA).
For 1 division, the voltage needed is 0.5 mV (because 2 divisions per mV).
Thus, the galvanometer resistance $G$ is given by:
$$ G = \frac{V_g}{I_g} = \frac{0.5\text{ mV}}{0.1\text{ mA}} = 5\,\Omega. $$
Step 3: Determine the Current for Full-Scale Deflection
• Full-scale deflection means 150 divisions.
• Since there are 10 divisions per mA, for 150 divisions:
$$ I_{\text{full-scale}} = \frac{150}{10} = 15\text{ mA}. $$
Step 4: Calculate the Total Required Resistance for 1 V per Division
• If each division is 1 V, then full-scale (150 divisions) corresponds to 150 V.
• Using Ohm’s law, the total resistance needed ($R_{\text{total}}$) to limit current to 15 mA when 150 V is applied is:
$$ R_{\text{total}} = \frac{V_{\text{full-scale}}}{I_{\text{full-scale}}}
= \frac{150\text{ V}}{15 \times 10^{-3}\text{ A}}
= 10,000\,\Omega. $$
Step 5: Subtract the Galvanometer Resistance to Find the Required Series Resistor
• The galvanometer itself has 5 Ω internal resistance.
• Hence, the series resistor $R$ must be:
$$ R = R_{\text{total}} - G = 10{,}000\,\Omega - 5\,\Omega = 9995\,\Omega. $$
Final Answer
The required series resistance is $9995\,\Omega.$
