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Question

(NONE) The total $e.m.f.$ of the battery $(E_{total})$ is $6 	imes 2\, V = 12\, V$.
The total internal resistance of the battery $(r_{total})$ is $6

Vedclass · Accepted Answer

$16A$

Vedclass · Answer

(NONE) The total $e.m.f.$ of the battery $(E_{total})$ is $6 	imes 2\, V = 12\, V$.
The total internal resistance of the battery $(r_{total})$ is $6 	imes 0.5\, \Omega = 3\, \Omega$.
The total resistance in the circuit $(R_{total})$ is the sum of the external resistor $(R = 10\, \Omega)$ and the internal resistance $(r_{total} = 3\, \Omega)$,which is $10 + 3 = 13\, \Omega$.
During charging,the effective voltage driving the current is the difference between the supply voltage $(V_{supply} = 220\, V)$ and the battery's $e.m.f.$ $(E_{total} = 12\, V)$.
Thus,the charging current $(I)$ is given by $I = \frac{V_{supply} - E_{total}}{R_{total}} = \frac{220 - 12}{13} = \frac{208}{13} = 16\, A$.

$A$ battery consisting of $6$ cells,each with an internal resistance of $0.5\, \Omega$ and an $e.m.f.$ of $2\, V$,is being charged by a $220\, V$ $D.C.$ mains using an external resistor of $10\, \Omega$. The charging current is: (in $, A$)

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