A milliammeter of range 10, mA and resistance | vedclass

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(C) The milliammeter has a full-scale deflection current $i_{g} = 10\, mA = 0.01\, A$ and resistance $R_{g} = 9\, \Omega$.When terminals $A$ and $B$ a

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$1\, A$

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(C) The milliammeter has a full-scale deflection current $i_{g} = 10\, mA = 0.01\, A$ and resistance $R_{g} = 9\, \Omega$.When terminals $A$ and $B$ are used,the current $I$ enters at $A$. The current $i_{g}$ flows through the milliammeter,and the remaining current $(I - i_{g})$ flows through the shunt resistor of $0.1\, \Omega$ connected between $A$ and $B$.Since the milliammeter and the shunt resistor are in parallel between $A$ and $B$,the potential difference across them is equal.$V_{AB} = i_{g} 	imes R_{g} = (I - i_{g}) 	imes R_{shunt}$$0.01\, A 	imes 9\, \Omega = (I - 0.01\, A) 	imes 0.1\, \Omega$$0.09 = 0.1I - 0.001$$0.1I = 0.091$$I = 0.91\, A = 910\, mA$.Wait,re-evaluating the circuit: The milliammeter is in series with the $0.9\, \Omega$ resistor,and this combination is in parallel with the $0.1\, \Omega$ resistor.$V_{AB} = i_{g}(R_{g} + 0.9\, \Omega) = (I - i_{g}) 	imes 0.1\, \Omega$$0.01(9 + 0.9) = (I - 0.01) 	imes 0.1$$0.01(9.9) = 0.1I - 0.001$$0.099 + 0.001 = 0.1I$$0.1 = 0.1I$$I = 1\, A$.

$A$ milliammeter of range $10\, mA$ and resistance $9\, \Omega$ is joined in a circuit as shown. The meter gives full-scale deflection for current $I$ when $A$ and $B$ are used as its terminals,i.e.,current enters at $A$ and leaves at $B$ ($C$ is left isolated). The value of $I$ is

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