In the following circuit diagram,E = 4,V, r = | vedclass

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(D) By analyzing the circuit using node labeling,we can identify the potential points. Let the negative terminal of the battery be at potential $0\,V$

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$1/4\, A$

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(D) By analyzing the circuit using node labeling,we can identify the potential points. Let the negative terminal of the battery be at potential $0\,V$. The positive terminal is at $4\,V$. Following the circuit,the three resistors $R$ are connected in parallel between the two nodes created by the battery and the shorting wire. Thus,the equivalent resistance of the three resistors $R$ in parallel is $R_{eq}' = R/3 = 45/3 = 15\,\Omega$. The total resistance of the circuit including the internal resistance $r$ is $R_{total} = R_{eq}' + r = 15 + 1 = 16\,\Omega$. The current $I$ flowing through the ammeter is given by Ohm's law: $I = E / R_{total} = 4 / 16 = 1/4\, A$.

In the following circuit diagram,$E = 4\,V, r = 1\,\Omega$ and $R = 45\, \Omega$,then the reading of the ammeter $A$ will be

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