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Question

A storage battery of emf $8.0\ V$ and internal resistance $0.5\, \Omega$ is being charged by a $120\ V$ dc supply using a series resistor of $15.5\, \Omega$. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?


Solution

Apply KVL in ABCD

$-8-0.{ 5 }_{ i }-15.{ 5 }_{ i }+120=0\\ 112={ 16 }_{ i }\\ i=\dfrac { 112 }{ 16 } A\\ i=7A$
By Going from point B to C:-
${ V }_{ B }-8-0.5\times 7-V_C=0$
${ V }_{ B }-{ V }_{C}=(8+3.5)V\\ $
Terminal voltage = 11.5V
$\rightarrow$ A series resistance in the charging circuit reduces the current value in the circuit. If resistor will not be there current will be extremely high.


891572_155547_ans_94ca8b6822534b98b8cdbe7f40eca312.JPG


Physics
NCERT
Standard XII

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