1. The time taken to discharge to 2 V is 2×10⁻⁹ s
2. The time taken to discharge to 1 V is 5×10⁻¹⁰ s
Energy stored in a capacitor
The energy stored in a capacitor is given by
E = ½CV²
But
E = Pt
Thus,
Pt = ½CV²
Where
- E is the energy
- C is the capacitor
- V is the voltage
- P is the power
- t is the time
With the formula (Pt = ½CV²), we can determine the time in each case. Detail below:
1. How to determine the time required to discharge to 2 V
Data obtained from the question include:
- Power (P) = 100 w
- Capacitor (C) = 0.10 μF = 1×10⁻⁷ F
- Voltage (V) = 2 V
- Time (t) = ?
Pt = ½CV²
100 × t = ½ × 1×10⁻⁷ × 2²
Divide both sides by 100
t = (½ × 1×10⁻⁷ × 2²) / 100
t = 2×10⁻⁹ s
Thus, the time required to discharge to 2 V is 2×10⁻⁹ s
2. How to determine the time required to discharge to 1 V
Data obtained from the question include:
- Power (P) = 100 w
- Capacitor (C) = 0.10 μF = 1×10⁻⁷ F
- Voltage (V) = 1 V
- Time (t) = ?
Pt = ½CV²
100 × t = ½ × 1×10⁻⁷ × 1²
Divide both sides by 100
t = (½ × 1×10⁻⁷ × 1²) / 100
t = 5×10⁻¹⁰ s
Thus, the time required to discharge to 1 V is 5×10⁻¹⁰ s
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