If all resistors are of 9, Omega,then the rea | vedclass

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(D) The circuit consists of a $9\, V$ battery connected in parallel with $9$ resistors,each of $9\, \Omega$.The ammeter is placed in series with one o

Vedclass · Accepted Answer

$9$

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(D) The circuit consists of a $9\, V$ battery connected in parallel with $9$ resistors,each of $9\, \Omega$.The ammeter is placed in series with one of the branches containing a $9\, \Omega$ resistor.According to Ohm's law,the current $I$ through a resistor is given by $I = \frac{V}{R}$.Since all resistors are connected in parallel to the $9\, V$ battery,the potential difference across each resistor is $9\, V$.Therefore,the current through the branch containing the ammeter is $I = \frac{9\, V}{9\, \Omega} = 1\, A$.Wait,looking at the circuit diagram,the ammeter is in series with only one branch. Thus,the reading is $1\, A$. However,checking the options provided,none match $1\, A$. Let's re-examine the circuit. If the ammeter is measuring the total current,there are $9$ resistors in parallel. Total resistance $R_{eq} = \frac{9}{9} = 1\, \Omega$. Total current $I = \frac{9\, V}{1\, \Omega} = 9\, A$. The ammeter is placed in the main line,so it measures the total current.Thus,the reading is $9\, A$.

If all resistors are of $9\, \Omega$,then the reading of the ammeter is .......... $A$.

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