A 50,V battery is connected across a 10,Omega | vedclass

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Question

(C) Given:Electromotive force $(E)$ = $50\,V$External resistance $(R)$ = $10\,\Omega$Current $(I)$ = $4.5\,A$Let the internal resistance of the batter

Vedclass · Accepted Answer

$1.1$

Vedclass · Answer

(C) Given:Electromotive force $(E)$ = $50\,V$External resistance $(R)$ = $10\,\Omega$Current $(I)$ = $4.5\,A$Let the internal resistance of the battery be $r$.According to Ohm's law for a circuit with an internal resistance:$I = \frac{E}{R + r}$Substituting the given values:$4.5 = \frac{50}{10 + r}$$4.5(10 + r) = 50$$45 + 4.5r = 50$$4.5r = 50 - 45$$4.5r = 5$$r = \frac{5}{4.5} = \frac{50}{45} = \frac{10}{9} \approx 1.11\,\Omega$.Thus,the internal resistance is approximately $1.1\,\Omega$.

$A$ $50\,V$ battery is connected across a $10\,\Omega$ resistor. The current is $4.5\,A$. The internal resistance of the battery is ............. $\Omega$.

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