Question

Two capacitors with capacitances of C₁ =0.3 μF and C₂=0.2 μF are

connected in series to an alternating current circuit with a voltage of 300

V and a frequency of 50 Hz. Find: (1) the intensity of the current in the

circuit, (2) the potential drop across the first and second capacitors.

Answer 1

(1) Let's first find the equivalent capacitance of two capacitors connected in series:

$$\dfrac{1}{C_{eq}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}},$$ $$C_{eq}=\dfrac{1}{\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}}=\dfrac{1}{\dfrac{1}{0.3\ \mu F}+\dfrac{1}{0.2\ \mu F}}=0.12\cdot10^{-6}\ F.$$

Then, we can find the capacitive reactance as follows:

$$X_C=\dfrac{1}{2\pi fC_{eq}}=\dfrac{1}{2\pi\cdot50\ Hz\cdot0.12\cdot10^{-6}\ F}=2.65\cdot10^4\ \Omega.$$

Finally, we can find the intensity of the current in the circuit:

$$I=\dfrac{V}{X_C}=\dfrac{300\ V}{2.65\cdot10^4\ \Omega}=1.1\cdot10^{-2}\ A.$$

(2) Let's first find the charge on the equivalent capacitor:

$$Q=C_{eq}\Delta V=0.12\cdot10^{-6}\ F\cdot300\ V=3.6\cdot10^{-5}\ C.$$

For the series connection of the capacitors, the magnitude of the charge will be the same, therefore, we can write:

$$Q_1=Q_2=Q.$$

Finally, we can find the potential drop across the first and second capacitors:

$$\Delta V_1=\dfrac{Q}{C_1}=\dfrac{3.6\cdot10^{-5}\ C}{0.3\cdot10^{-6}\ F}=120\ V,$$ $$\Delta V_2=\dfrac{Q}{C_2}=\dfrac{3.6\cdot10^{-5}\ C}{0.2\cdot10^{-6}\ F}=180\ V.$$