The ammeter reading in the following circuit | vedclass

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(C) From the circuit diagram,we observe that there are three $6 \Omega$ resistors connected in parallel with each other. Let their equivalent resistan

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(C) From the circuit diagram,we observe that there are three $6 \Omega$ resistors connected in parallel with each other. Let their equivalent resistance be $R_p$.$\frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \Omega^{-1}$,so $R_p = 2 \Omega$.This parallel combination is in series with the $2 \Omega$ resistor present in the branch containing the battery and the ammeter.Total resistance of the circuit $R_{eq} = R_p + 2 \Omega = 2 \Omega + 2 \Omega = 4 \Omega$.The voltage of the battery is $V = 2 	ext{ V}$.Using Ohm's law,the current $I$ measured by the ammeter is $I = \frac{V}{R_{eq}} = \frac{2 	ext{ V}}{4 \Omega} = 0.5 	ext{ A}$.

The ammeter reading in the following circuit will be .............. $A$.

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