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A storage battery of emf $8.0 \mathrm{~V}$ and internal resistance
$0.5 \Omega$ is being charged by a 120 V D.C. 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?
$0.5 \Omega$ is being charged by a 120 V D.C. 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:
2965 Upvotes
Verified Answer
The total circuit resistance becomes
$$
15.5+0.5=16 \Omega \text {. }
$$
The total e.m.f. of circuit will be $120-8=112 \mathrm{~V}$ (Battery emf opposing supply emf)
Charging current becomes $\frac{112}{16}=7.0 \mathrm{~A}$.
For battery being charged, terminal voltage,
$$
\mathrm{V}=\mathrm{E}+\mathrm{Ir}=8.0+7 \times 0.5=11.5 \mathrm{~V} \text {. }
$$
A series resistor is used to reduce charging current to a proper value.
$$
15.5+0.5=16 \Omega \text {. }
$$
The total e.m.f. of circuit will be $120-8=112 \mathrm{~V}$ (Battery emf opposing supply emf)
Charging current becomes $\frac{112}{16}=7.0 \mathrm{~A}$.
For battery being charged, terminal voltage,
$$
\mathrm{V}=\mathrm{E}+\mathrm{Ir}=8.0+7 \times 0.5=11.5 \mathrm{~V} \text {. }
$$
A series resistor is used to reduce charging current to a proper value.
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