Question

Find the heat produced in the capacitors on closing the switch S.

Answer 1

Hint: This question demands a qualitative approach rather than a mathematical treatment. So instead of diving right into calculations, intuitively deduce the behaviour of the circuit before and after closing the switch. Recall that a capacitor stores charge when connected to a voltage source. Think of how and if this would change when the switch is closed, given that the voltage source still remains in place.

Complete answer:
We know that capacitors are devices that are capable of storing electrical energy. A capacitor consists of two conductive plates separated by a non-conducting region such as vacuum or insulating material. When connected to a voltage source, the charges (current) reaching one plate will exert an attractive force on the opposite charges within the other plate causing an electric field to form between the two plates. The magnitude of this field determines the charge stored by the capacitor.
Let us now look at the circuit given to us.
We have an open switch in the circuit, and we have a 20V voltage source that is directly connected to a $4\;\mu F$ capacitor. Without the voltage source, the charge stored in the capacitor will be 0. But since we have a voltage source, it becomes responsible for conveying charges to the capacitor, which is called transient current. This transient current continues to flow between the voltage source and the capacitor until the capacitor gets charged to the value of the applied voltage. The capacitor thus acts like a storage device, holding this charge indefinitely as long as the supply voltage is present. When the capacitor reaches the point where it has more or less acquired and stored charge equivalent to dc voltage, it is said to be in a steady state. At this point, there is no potential difference between the voltage source and the capacitor, and no current flows through the capacitor to the circuit and the capacitor acts like an open circuit.
Now, when the switch is closed, there still won’t be any current flow across the rest of the circuit. This is because the $4\;\mu F$ capacitor continues to act as an open circuit since the voltage source is still in place. Since the voltage source is cut off from the rest of the circuit, there won’t be any current flowing through any parts of the circuit, which leaves no room for any heat to be produced. The capacitor does not discharge since it maintains a constant potential with the voltage source and does not allow it to set up any current in the rest of the circuit.
The capacitor will have an energy of $E = \dfrac{1}{2}CV^2 = \dfrac{1}{2} \times 4 \times 10^{-6} \times 20^2 = 0.8\;\mu J$ stored in any case, and ideally none of it is lost to heat production if it has no internal resistance or does not discharge to a finite resistance.
Thus, the heat produced in the capacitors upon closing the switch will be zero.

Note:
It is important to understand that a capacitor behaves as an open circuit for dc current but acts like a high pass filter (allows high frequency oscillations) for ac current. This can be understood by capacitive reactance, which is defined as the resistance offered to the flow of current by a capacitor. It is given as:
$X_{C} = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$
We know that dc current does not oscillate, so $f = 0 \Rightarrow X_C = \dfrac{1}{0} = \infty$,
Whereas for ac high frequencies, we have $X_C \rightarrow 0$
This means that a capacitor offers infinite resistance to the flow of dc, so dc cannot pass through the capacitor, but a capacitor offers only a small resistance to the flow of ac. Thus, our discussion in the question is valid only for dc sources.