Question

A 0.25μF capacitor is charged to 100 V . It is then connected in series with a 45Ω resistor and a 120 Ω resistor and allowed to discharge completely. How much energy is dissipated by the 45Ω resistor?

Answer 1

Answer 2

Calculation :

1. Find the Total Energy Stored in the Capacitor Initially

The capacitor starts with energy:

$$E_{\text{total}}=\frac{1}{2}C{V}^2=\frac{1}{2}(0.25\times 10^{-6}\text{ F})(100\text{ V}{)}^2=0.00125\text{ J}$$

 

2. Determine the Total Resistance in the Circuit

The resistors are in series, so:

$$R_{\text{total}}=450\mathrm{\Omega }+120\mathrm{\Omega }=570\mathrm{\Omega }$$

3. Calculate the Fraction of Energy Dissipated by the 450 Ω Resistor

Energy splits proportionally to resistance in series. The 450 Ω resistor gets:

$$E_{450}=\left(\frac{450}{570}\right)\times E_{\text{total}}=\left(\frac{450}{570}\right)\times 0.00125\text{ J}=0.00113\text{ J}$$

 

Key Notes:

• The 120 Ω resistor dissipates the remaining 0.00012 J.

• All energy is accounted for: $0.00113+0.00012=0.00125\text{ J}$$0.00125\text{ J}$.

 Answer:

The 450 Ω resistor dissipates 0.00113 J (1.13 mJ) of energy during the capacitor's discharge.