In this section we'll study on the effective capacitance of the capacitors in the circuit.
Firstly, the effective capacitance connected in series in the circuit shown below,
The effective capacitance '$C$' is given by the following expression, $$\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}...$$
The expression shown above can be derived by adding the potential across each of the capacitance across the capacitor. We know from kirchoff law and also from the fact that each capacitor has the same amount of charge flowing through it, an equation can be written for the capacitance of a capacitors in series that,
$$V=V_{1}+V_{2}....(1)$$ And also from the relation,$$V=\frac{Q}{C}....(2)$$so substituting $..(2)$ in $..(1)$ we get, $$\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}...$$
Secondly, the capacitor in the parallel arrangement as shown below,
The effective capacitance '$C$' is given by the following expression, $$C=C_{1}+C_{2}...$$
The expression shown above can be derived by using the principle of the conservation of the charge and using the fact that the voltage is the same for each capacitor in parallel, $$Q=Q_{1}+Q_{2}....(3)$$ substituting $..(2)$ in $..(3)$ we get, $$C=C_{1}+C_{2}...$$
Here's an easy to understand another basics on capacitors,
Firstly, the effective capacitance connected in series in the circuit shown below,
The expression shown above can be derived by adding the potential across each of the capacitance across the capacitor. We know from kirchoff law and also from the fact that each capacitor has the same amount of charge flowing through it, an equation can be written for the capacitance of a capacitors in series that,
$$V=V_{1}+V_{2}....(1)$$ And also from the relation,$$V=\frac{Q}{C}....(2)$$so substituting $..(2)$ in $..(1)$ we get, $$\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}...$$
Secondly, the capacitor in the parallel arrangement as shown below,
The expression shown above can be derived by using the principle of the conservation of the charge and using the fact that the voltage is the same for each capacitor in parallel, $$Q=Q_{1}+Q_{2}....(3)$$ substituting $..(2)$ in $..(3)$ we get, $$C=C_{1}+C_{2}...$$
Here's an easy to understand another basics on capacitors,
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