First slide

Electrical measuring instruments

Question

Two meters of voltage range 20.0 V and 30.0 V have to be constructed with a galvanometer. The resistance connected in series with the galvanometer is $1680 Ω$ for the 20.0 V range and $2930 Ω$ for the 30.0 V range. The resistance of the galvanometer and the full -scale current are respectively.

Moderate

A

$320 Ω and 8 mA$
B

$70 Ω and 10 mA$
C

$820 Ω and 10 mA$
D

$820 Ω and 8 mA$

Solution

Given that
$\begin{aligned} v = 20.0 V for R_{1} = 1680 Ω \\ v = 30.0 V for R_{2} = 2930 Ω \end{aligned}$
Let coil resistance = G
If full scale current is $I_{g}$
$\begin{array}{l} I_{g} (G + R_{1}) = 20 \\ \Rightarrow I_{g} (G + 1680) = 20 \dots . . (1) \\ And I_{g} (G + 2930) = 30 \dots . . (2) \\ Dividing (1) & (2) \\ \frac{G + 1680}{G + 2930} = \frac{20}{30} \\ \Rightarrow 3 (G + 1680) = 2 (G + 2930) \\ \Rightarrow G = 5860 - 5040 \\ ∴ G = 820 Ω \end{array}$
$\begin{array}{l} From (1) \\ I_{g} = \frac{20}{820 + 1680} \\ = 8 \times 10^{- 3} A \\ ∴ I_{g} = 8 mA \end{array}$

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