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Step-by-Step Solution
Step 1: Understand the Given Quantities
• The current sensitivity of the galvanometer is given as 5 div/mA. This means that the galvanometer shows 5 scale divisions for every 1 mA of current passing through it.
• The voltage sensitivity of the galvanometer is 20 div/V. This means that the galvanometer shows 20 scale divisions for every 1 V applied across it.
Step 2: Write the Relevant Relation
Let $I_s$ be the current sensitivity and $V_s$ be the voltage sensitivity. For a moving coil galvanometer of resistance $R_G$, we have:
$R_G = \frac{I_s}{V_s}$
This relationship comes from the expression for current sensitivity $I_s$ and voltage sensitivity $V_s$:
• $I_s$ (in div/A) tells us how many divisions are seen per ampere of current.
• $V_s$ (in div/V) tells us how many divisions are seen per volt of applied potential.
By definition, $V_s = \frac{I_s}{R_G}.$ Hence, rearranging gives $R_G = \frac{I_s}{V_s}.$
Step 3: Substitute the Numerical Values
• Current sensitivity, $I_s = 5\text{ div/mA} = 5\, \text{div} \,/\, (1 \times 10^{-3} \text{A}) = 5000\,\text{div/A}.$
• Voltage sensitivity, $V_s = 20\text{ div/V}.$
Thus,
$R_G = \frac{I_s}{V_s}
= \frac{5000\,\text{div/A}}{20\,\text{div/V}}
= \frac{5000}{20}
= 250\,\Omega.$
Step 4: Conclude the Answer
Therefore, the resistance of the galvanometer is $250\,\Omega.$
