Question
Show that only half the total energy drawn from a battery in charging an $R C$ circuit ends up stored in the capacitor. (Hint:What happens to the rest? You'll need to integrate.)
Step 1
We have a series RC circuit consisting of a resistor \( R \) and a capacitor \( C \) connected to a battery of voltage \( V \). When the switch is closed, the capacitor starts charging through the resistor. Show more…
Show all steps
Your feedback will help us improve your experience
Suman Saurav Thakur and 81 other Physics 102 Electricity and Magnetism educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The current in a charging capacitor is given by Eq. (26.13) (a) The instantaneous power supplied by the battery is $mathcal{E} i$. Integrate this to find the total energy supplied by the battery. (b) The instantaneous power dissipated in the resistor is $i^{2} R$. Integrate this to find the total energy dissipated in the resistor. (c) Find the final energy stored in the capacitor, and show that this equals the total energy supplied by the battery less the energy dissipated in the resistor, as obtained in parts (a) and (b). (d) What fraction of the energy supplied by the battery is stored in the capacitor? How does this fraction depend on $R ?$
When capacitor of capacitance $C$ is charged by a source of voltage $V,$ the power expended at time $t$ is $$ P(t)=\frac{V^{2}}{R}\left(e^{-t / R C}-e^{-2 t / R C}\right) $$ where $R$ is the resistance in the circuit. The total energy stored in the capacitor is $$ W=\int_{0}^{\infty} P(t) d t $$ Show that $W=\frac{1}{2} C V^{2}$
TECHNIQUES OF INTEGRATION
Improper Integrals
When a capacitor of capacitance $C$ is charged by a source of voltage $V,$ the power expended at time $t$ is $$P(t)=\frac{V^{2}}{R}\left(e^{-t / R C}-e^{-2 t / R C}\right)$$ where $R$ is the resistance in the circuit. The total energy stored in the capacitor is $$W=\int_{0}^{\infty} P(t) d t$$ Show that $W=\frac{1}{2} C V^{2}$
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD