Two batteries of emf's 6.20 V and 12.45 V and internal resistances 0.0...
Terminal Potential Differences of Batteries during Charging
Given Data:
- EMF of first battery, E1 = 6.20 V
- EMF of second battery, E2 = 12.45 V
- Internal resistance of first battery, r1 = 0.01 Ω
- Internal resistance of second battery, r2 = 0.03 Ω
- Current supplied by battery charger, I = 20 A
Explanation:
When the batteries are being charged, the current supplied by the battery charger flows through the internal resistance and the battery. As a result, there will be a voltage drop across the internal resistance of the battery, which reduces the terminal potential difference of the battery.
The terminal potential difference of the battery during charging can be calculated using the following formula:
V = E - Ir
where V is the terminal potential difference of the battery, E is the EMF of the battery, I is the current supplied by the battery charger, and r is the internal resistance of the battery.
Calculation:
Let us calculate the terminal potential difference of each battery during charging using the above formula.
For the first battery:
V
1 = E
1 - Ir
1 = 6.20 V - (20 A x 0.01 Ω)
= 5.80 V
Therefore, the terminal potential difference of the first battery during charging is 5.80 V.
For the second battery:
V
2 = E
2 - Ir
2 = 12.45 V - (20 A x 0.03 Ω)
= 11.85 V
Therefore, the terminal potential difference of the second battery during charging is 11.85 V.
Conclusion:
During charging, the terminal potential difference of a battery is reduced due to the voltage drop across the internal resistance of the battery. In this case, the terminal potential differences of the first and second batteries are 5.80 V and 11.85 V respectively.