(B) The charging current $i$ is given by the formula $i = \frac{V_{supply} - V_{battery}}{R_{ext} + r_{int}}$.Substituting the given values: $i = \fra

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(B) The charging current $i$ is given by the formula $i = \frac{V_{supply} - V_{battery}}{R_{ext} + r_{int}}$.Substituting the given values: $i = \frac{24 - 12}{2 + 1} = \frac{12}{3} = 4\, A$.The energy stored in the battery is given by $E = V_{battery} \times i \times t$.Given $E = 360\, Wh$, $V_{battery} = 12\, V$, and $i = 4\, A$.$360 = 12 \times 4 \times t$.$360 = 48 \times t$.$t = \frac{360}{48} = 7.5\, hours$.

Class 12 Physics Current Electricity Circuit Solving for current and Voltage

Difficult

$A$ $12\, V$ lead accumulator is being charged using a $24\, V$ supply with an external resistance of $2\,\Omega$. The internal resistance of the accumulator is $1\,\Omega$. Find the time in which it will store $360\, Wh$ of energy.

A
$1$
B
$7.5$
C
$10$
D
None of these

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Current Electricity

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Circuit Solving for current and Voltage

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DifficultJEE MAIN 2024

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$A$ $12\, V$ lead accumulator is being charged using a $24\, V$ supply with an external resistance of $2\,\Omega$. The internal resistance of the accumulator is $1\,\Omega$. Find the time in which it will store $360\, Wh$ of energy.

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$A$ cell of internal resistance $r$ is connected across an external resistance $n r$. Then the ratio of the terminal voltage to the emf of the cell is

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