A new flashlight cell of e.m.f. 1.5, V gives | vedclass

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(B) Given: $e.m.f.$ $(E)$ = $1.5\, V$,Current $(I)$ = $15\, A$,External resistance $(R)$ = $0.04\,\Omega$.Using the formula for current in a circuit w

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$0.06$

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(B) Given: $e.m.f.$ $(E)$ = $1.5\, V$,Current $(I)$ = $15\, A$,External resistance $(R)$ = $0.04\,\Omega$.Using the formula for current in a circuit with internal resistance $(r)$: $I = \frac{E}{R + r}$.Substituting the values: $15 = \frac{1.5}{0.04 + r}$.Rearranging the equation: $0.04 + r = \frac{1.5}{15}$.$0.04 + r = 0.1$.$r = 0.1 - 0.04 = 0.06\,\Omega$.Therefore,the internal resistance of the cell is $0.06\,\Omega$.

$A$ new flashlight cell of $e.m.f.$ $1.5\, V$ gives a current of $15\, A$,when connected directly to an ammeter of resistance $0.04\,\Omega$. The internal resistance of the cell is ........... $\Omega$.

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