An ammeter with internal resistance 90,Omega | vedclass

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Question

(B) Let the electromotive force $(EMF)$ of the battery be $E$. When the ammeter is connected in series,the total resistance of the circuit is $R_{tota

Vedclass · Accepted Answer

Greater than $1.85\,A$

Vedclass · Answer

(B) Let the electromotive force $(EMF)$ of the battery be $E$. When the ammeter is connected in series,the total resistance of the circuit is $R_{total} = R_1 + R_2 + R_A = 700\,\Omega + 410\,\Omega + 90\,\Omega = 1200\,\Omega$. The current measured by the ammeter is $I_{measured} = E / R_{total} = 1.85\,A$. Therefore,$E = 1.85\,A 	imes 1200\,\Omega = 2220\,V$. When the ammeter is removed,the actual resistance of the circuit is $R_{actual} = R_1 + R_2 = 700\,\Omega + 410\,\Omega = 1110\,\Omega$. The actual current is $I_{actual} = E / R_{actual} = 2220\,V / 1110\,\Omega = 2.0\,A$. Since $2.0\,A > 1.85\,A$,the actual current is greater than the measured current.

An ammeter with internal resistance $90\,\Omega$ reads $1.85\,A$ when connected in a circuit containing a battery and two resistors $700\,\Omega$ and $410\,\Omega$ in series. The actual current in the circuit (without the ammeter) will be:

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