The reading of the ammeter in the following f | vedclass

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(C) From the circuit diagram, the resistors $2 \Omega$, $3 \Omega$, and $6 \Omega$ are connected in parallel.Let their equivalent resistance be $R_p$.

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(C) From the circuit diagram, the resistors $2 \Omega$, $3 \Omega$, and $6 \Omega$ are connected in parallel.Let their equivalent resistance be $R_p$. Then, $\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3+2+1}{6} = 1 \Omega^{-1}$.Thus, $R_p = 1 \Omega$.This parallel combination is in series with the $4 \Omega$ resistor and the $2 	ext{ V}$ battery.Therefore, the total equivalent resistance of the circuit is $R_{	ext{eq}} = R_p + 4 \Omega = 1 \Omega + 4 \Omega = 5 \Omega$.The current $I$ measured by the ammeter is given by Ohm's law: $I = \frac{V}{R_{	ext{eq}}} = \frac{2 	ext{ V}}{5 \Omega} = 0.4 	ext{ A}$.

The reading of the ammeter in the following figure will be: (in $\text{A}$)

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