An electrical meter of internal resistance 20Ω gives a full scale deflection when one milliampere current flows through it. The maximum current, that can be measured by using three resistors of resistance 12Ω each, in milliamperes is:

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answer is C.

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Detailed Solution

$\begin{array}{l} G = 20 Ω \Rightarrow i_{g} = 10^{- 3} A; R_{eq} = \frac{R}{3} = \frac{12}{3} = 4 \\ S = \frac{G}{(i / i_{g} - 1)} \Rightarrow 4 = \frac{20}{(I / 10^{- 3} - 1)} \\ \frac{I}{10^{- 3}} - 1 = 5 I (10^{+ 3}) = 6 \Rightarrow I = 6 mA \end{array}$

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