A circuit contains an ammeter,a battery of 30 | vedclass

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(C) The circuit is shown in the figure.First,calculate the effective resistance of the ammeter $(R_A)$. Since the coil resistance $(480\,\Omega)$ and

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$0.5$

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(C) The circuit is shown in the figure.First,calculate the effective resistance of the ammeter $(R_A)$. Since the coil resistance $(480\,\Omega)$ and the shunt resistance $(20\,\Omega)$ are in parallel,the equivalent resistance is:$R_A = \frac{480 	imes 20}{480 + 20} = \frac{9600}{500} = 19.2\,\Omega$Since the ammeter is connected in series with the $40.8\,\Omega$ resistor,the total resistance $(R_{total})$ of the circuit is:$R_{total} = 40.8\,\Omega + 19.2\,\Omega = 60\,\Omega$Using Ohm's law $(I = V/R)$,the total current flowing through the circuit is:$I = \frac{30\,V}{60\,\Omega} = 0.5\,A$Thus,the reading in the ammeter will be $0.5\,A$.

$A$ circuit contains an ammeter,a battery of $30\,V$ and a resistance of $40.8\,\Omega$ all connected in series. If the ammeter has a coil of resistance $480\,\Omega$ and a shunt of $20\,\Omega$,the reading in the ammeter will be .................. $A$.

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