If the charge on a capacitor is increased by 2 C, the energy stored in it increases by 44%. The original charge on the capacitor is (in C)
- 10
- 20
- 30
- 40
The Correct Option is A
Solution and Explanation
Let initially the charge is \(q\) so
\(\frac{1}{2} \frac{q^2}{C} = U_i\)
And
\(\frac{1}{2} \frac{(q+2)^2}{C} = U_f\)
Given
\(\frac{U_f -Ui}{Ui}\times 100 = 44\)
\(\frac{(q+2)^2-q^2}{q} = 0.44\)
⇒ \(q = 10 \;C\)
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Concepts Used:
Combination of Capacitors
The total capacitance of this equivalent single capacitor depends both on the individual capacitors and how they are connected. There are two simple and common types of connections, called series and parallel, for which we can easily calculate the total capacitance.
Read Also: Combination of Capacitors
Series capacitors
When one terminal of a capacitor is connected to the terminal of another capacitors , called series combination of capacitors.
Capacitors in Parallel
Capacitors can be connected in two types which are in series and in parallel. If capacitors are connected one after the other in the form of a chain then it is in series. In series, the capacitance is less.
When the capacitors are connected between two common points they are called to be connected in parallel.
When the plates are connected in parallel the size of the plates gets doubled, because of that the capacitance is doubled. So in a parallel combination of capacitors, we get more capacitance.
Read More: Types of Capacitors