Given Data:
- Electromotive force (emf) of the battery, $E = 8.0 \text{ V}$
- Internal resistance of the battery, $r = 0.5 \text{ } \Omega$
- DC supply voltage, $V_{supply} = 120 \text{ V}$
- Series resistance, $R = 15.5 \text{ } \Omega$
Part 1: Terminal Voltage of the Battery
To find the terminal voltage of the battery during charging, we first need to calculate the charging current flowing through the circuit.
1. Find the Effective Voltage:
When charging, the battery's emf (8.0 V) opposes the DC supply's voltage (120 V). Therefore, the net or effective voltage that drives the current in the circuit is the difference between the supply voltage and the battery's emf.
$V{eff} = V{supply} - E$
$V_{eff} = 120 \text{ V} - 8.0 \text{ V} = 112 \text{ V}$
2. Find the Total Resistance:
The total resistance in the circuit is the sum of the series resistor and the battery's internal resistance, as they are connected in series.
$R{total} = R + r$
$R{total} = 15.5 \text{ } \Omega + 0.5 \text{ } \Omega = 16.0 \text{ } \Omega$
3. Calculate the Charging Current (I):
Using Ohm's Law for the entire circuit:
$I = \frac{V{eff}}{R{total}}$
$I = \frac{112 \text{ V}}{16.0 \text{ } \Omega} = 7.0 \text{ A}$
4. Calculate the Terminal Voltage (V_T):
The terminal voltage across the battery during charging is the sum of its emf and the voltage drop across its internal resistance. This is because the external supply must overcome both the battery's emf and its internal resistance to push current into it.
The formula is: $V_T = E + Ir$
$V_T = 8.0 \text{ V} + (7.0 \text{ A} \times 0.5 \text{ } \Omega)$
$V_T = 8.0 \text{ V} + 3.5 \text{ V}$
$V_T = 11.5 \text{ V}$
Answer: The terminal voltage of the battery during charging is 11.5 V.
Part 2: Purpose of the Series Resistor
The primary purpose of the series resistor in a charging circuit is to limit the charging current to a safe level.
Explanation:
If the series resistor were not in the circuit, the only resistance would be the battery's small internal resistance ($0.5 \text{ } \Omega$). Let's see what would happen:
- Without the resistor, the charging current would be:
$I = \frac{V_{eff}}{r} = \frac{112 \text{ V}}{0.5 \text{ } \Omega} = 224 \text{ A}$
This extremely high current of 224 A would cause severe problems:
- Overheating: The power dissipated as heat inside the battery ($P = I^2r$) would be enormous. This would cause the battery to overheat rapidly, potentially boiling the electrolyte and causing permanent damage.
- Damage to the Battery: Excessive current can damage the battery's internal plates, reduce its lifespan, and in extreme cases, lead to a risk of explosion or fire.
- Damage to the Power Supply: The 120 V DC supply might not be designed to deliver such a high current and could be damaged or its protective circuits would trip.
By adding the 15.5 $\Omega$ series resistor, the current is controlled and reduced to a much safer and more manageable 7.0 A, protecting both the battery and the power supply from damage.
