First slide

Moving coil Galvanometer

Question

In a galvanometer, there is deflection of 5 divisions per $μ$ A current. The resistance of galvanometers is 40 ohm. If shunt of 2 ohm is connected and if there are 50 divisions all over the scale of galvanometer, then the maximum current that can be measured by it is

Moderate

A

2.1 mA
B

0.21 mA
C

0.021 mA
D

5.25 mA

Solution

$\begin{array}{rcl} s & = & \frac{G}{n - 1} where G is the galvanometer resistance \\ and n is the amplification factor \\ 2 & = & \frac{40}{n - 1}; n = 21 \end{array}$
$\begin{array}{rcl} \begin{array}{lcl} before \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} connecting \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} shunt \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} it \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} measures \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} a \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} current \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} of \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} 50 \end{array} \begin{array}{rcl} \times \end{array} \begin{array}{rcl} \frac{1}{5} \end{array} & = & \begin{array}{rcl} 10 \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} μA \end{array} \begin{array}{rcl} \end{array} \\ \begin{array}{rcl} now \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} it \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} measures \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} 21 \end{array} \begin{array}{rcl} \times \end{array} \begin{array}{rcl} 10 \end{array} & = & \begin{array}{rcl} 210 \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} μA \end{array} = \begin{array}{rcl} 0 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 21 \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} mA \end{array} \end{array}$

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