During a battery charging process,During a battery charging process, the emf of the battery increases from 10 V to 12 V linearly over a period of 6 hours. The charging voltage is 15V. A series connected resistor is continuously adjusted to maintain a constant current of 1A. What is the energy wasted in the system assuming ideal charging source and ideal battery.
During a battery charging process, the emf of the battery increases from 10 V to 12 V linearly over a period of 6 hours. The charging voltage is 15V. A series connected resistor is continuously adjusted to maintain a constant current of 1A. What is the energy wasted in the system assuming ideal charging source and ideal battery.
Right Answer is:
18 Wh
SOLUTION
Voltage across the Resistor:
- The charging voltage is 15V.
- The battery voltage increases linearly from 10V to 12V over 6 hours.
- To maintain a constant current of 1A, the voltage drop across the resistor must be constant.
Therefore, the voltage across the resistor is 15V – 12V = 3V.
- Resistance of the Resistor:
Using Ohm’s law, we can calculate the resistance:
R = V / I
= 3V / 1A = 3Ω
- Energy Dissipated in the Resistor:
The energy dissipated in a resistor is given by the formula:
Energy = Power × Time = (I2)R × t
where:
- I = Current = 1A
- R = Resistance = 3Ω
- t = Time = 6 hours
So, the energy wasted is:
Energy = (12) × 3 × 6 = 18 Wh