The reading of the ammeter (A) in steady stat | vedclass

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(B) In steady state, the capacitor acts as an open circuit, so no current flows through the branch containing the capacitor.The circuit simplifies to

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) In steady state, the capacitor acts as an open circuit, so no current flows through the branch containing the capacitor.The circuit simplifies to a $10 \ V$ battery connected to a $1 \ \Omega$ resistor in series with a parallel combination of two $8 \ \Omega$ resistors.The equivalent resistance of the two $8 \ \Omega$ resistors in parallel is $R_p = (8 \ \Omega 	imes 8 \ \Omega) / (8 \ \Omega + 8 \ \Omega) = 4 \ \Omega$.The total resistance of the circuit is $R_{eq} = 1 \ \Omega + 4 \ \Omega = 5 \ \Omega$.The total current drawn from the battery is $I = V / R_{eq} = 10 \ V / 5 \ \Omega = 2 \ A$.The ammeter is placed in the return path of the circuit. Since the capacitor branch is open, the current $I = 2 \ A$ flows through the $1 \ \Omega$ resistor and then splits equally into the two $8 \ \Omega$ resistors. The ammeter measures the current flowing through the right-hand side of the circuit, which is the current through the rightmost $8 \ \Omega$ resistor. Since the current $I = 2 \ A$ splits equally into the two $8 \ \Omega$ resistors, the current through each is $I' = I / 2 = 2 \ A / 2 = 1 \ A$.Therefore, the reading of the ammeter is $1 \ A$.

The reading of the ammeter $(A)$ in steady state in the following circuit (assuming negligible internal resistance of the ammeter) is . . . . . . $A$.

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