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?
Exercise
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Emf of the storage battery $, E = 8.0 V$
Internal resistance of the battery, $\text{r}=0.5\ \Omega$
$DC$ supply voltage $, V = 120 V$
Resistance of the resistor $, \text{R}=15.5\ \Omega$
Effective voltage in the circuit $= V^1$
$R$ is connected to the storage battery in series.
Hence, it can be written as
$V^1 = V - E$
$V^1 = 120 - 8 = 112 V$
Current flowing in the circuit $= I,$ which is given by the relation,
$\text{I}=\frac{\text{V}^1}{\text{R}+\text{r}}$
$=\frac{112}{15.5+5}=\frac{112}{16}=7\text{A}$
Voltage across resistor R given by the product $, IR = 7 \times 15.5 = 108.5 V$
$DC$ supply voltage $=$ Terminal voltage of battery $+$ Voltage drop across $R$
Terminal voltage of battery $= 120 - 108.5 = 11.5 V$
A series resistor in a charging circuit Itmtts the current drawn from the external source.
The current will be extremely high in its absence. This is very dangerous.
Internal resistance of the battery, $\text{r}=0.5\ \Omega$
$DC$ supply voltage $, V = 120 V$
Resistance of the resistor $, \text{R}=15.5\ \Omega$
Effective voltage in the circuit $= V^1$
$R$ is connected to the storage battery in series.
Hence, it can be written as
$V^1 = V - E$
$V^1 = 120 - 8 = 112 V$
Current flowing in the circuit $= I,$ which is given by the relation,
$\text{I}=\frac{\text{V}^1}{\text{R}+\text{r}}$
$=\frac{112}{15.5+5}=\frac{112}{16}=7\text{A}$
Voltage across resistor R given by the product $, IR = 7 \times 15.5 = 108.5 V$
$DC$ supply voltage $=$ Terminal voltage of battery $+$ Voltage drop across $R$
Terminal voltage of battery $= 120 - 108.5 = 11.5 V$
A series resistor in a charging circuit Itmtts the current drawn from the external source.
The current will be extremely high in its absence. This is very dangerous.
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