A cell of emf 1.8 V gives a current of 17 A | vedclass

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(A) Let the internal resistance of the cell be $r$. The ammeter is connected in series with the cell,so the total resistance of the circuit is $(R + r

Vedclass · Accepted Answer

$0.046$

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(A) Let the internal resistance of the cell be $r$. The ammeter is connected in series with the cell,so the total resistance of the circuit is $(R + r)$,where $R = 0.06 \ \Omega$ is the resistance of the ammeter.According to Ohm's law for a complete circuit,the emf $E$ is given by $E = I(R + r)$.Given $E = 1.8 \ V$,$I = 17 \ A$,and $R = 0.06 \ \Omega$.Substituting the values: $1.8 = 17(0.06 + r)$.$1.8 = 1.02 + 17r$.$17r = 1.8 - 1.02 = 0.78$.$r = \frac{0.78}{17} \approx 0.04588 \ \Omega$.Rounding to three decimal places,we get $r \approx 0.046 \ \Omega$.

$A$ cell of emf $1.8 \ V$ gives a current of $17 \ A$ when directly connected to an ammeter of resistance $0.06 \ \Omega$. The internal resistance of the cell is: (in $\Omega$)

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