Question

A capacitor of capacitance 2.0 mu F is charged by connecting it to a battery of emf 5V .The capacitor is now disconnected and reconnected to the same battery with the polarity reversed.The heat developed in the connecting wires is​

Answer 1

The heat developed in the connecting wire = 2 × 10⁻⁵ J

Explanation:

Given:

The capacitance of the capacitor = 2 μF

EMF of the battery = 5V

To find out:

The heat developed in the connecting wire when the capacitor is connected to the same battery with its polarity reversed

Solution:

Given that

C = 2 μF

ε = 5V

When the capacitor C is connected to the battery of emf ε, the charge on the capacitor plates will be +Cε and -Cε

When the capacitor is disconnected and again connected with the battery of same emf but polarity reversed, the charge now will be -Cε and +Cε

Thus, the total charge flown from the battery

= Cε - (-Cε)

= 2Cε

The Work done will be

W = qV

W = 2Cε × ε

or, W = 2Cε²

        = 2 × 2 × 10⁻⁶ × 5 J

        = 2 × 10⁻⁵ J

The potential energy stored in the capacitor initially

= $\frac{1}{2}CV^2$

= $\frac{1}{2}C\epsilon^2$

The potential energy stored in the capacitor finally

= $\frac{1}{2}C\epsilon^2$

Change in Potential Energy $\Delta U=0$

Therefore, the heat dissipated will be equal to the work done

= 2 × 10⁻⁵ J

Hope this answer is helpful.

Know More:

Q: A capacitor of capacitance 'C' is initially charged to a potential difference 'V' volt. now it is connected to a battery of 2V volt with opposite polarity . the ratio of heat generated to the final energy stored in the capacitor will be?

Click Here: https://brainly.in/question/4299529

Answer 2

Answer:

100uj

Explanation:

hopefully this ans will help u