(a)(i) What is meant by the root-mean-square value of an alternating current? (ii) Define impedance...
(a)(i) What is meant by the root-mean-square value of an alternating current? (ii) Define impedance of an alternating current circuit.
(b) An electrical device rated 120 V, 60 W is opened on a 240 V, 50Hz mains supply. The circuit has a capacitor connected in series with ihe electrical device and the supply. Calculate the capacitance of the capacitor. [π=3.142].
(c)(i) Define the capacitance of a capacitor.
(ii)
.The circuit diagram below illustrates two capacitors of capacitance C\(_1\) and C\(_2\) connected in series across a 2V source.
(i)Obtain an expression for the total capacitance in terms of C\(_2\). 2 mm 5 n (ii) Calculate the potential difference across each capacitor.
Explanation
(a)G) Root-mean-square value of an alternating Current 1s the value of a.c. current that has the same heating effect as d.c. current.
(ii) Impedance is the effective resistance of an electric circuit arising from the combined effects of ohmic resistance, capacitance and inductive reactance.
(b) Current in circuit I = 60/120 = 0.5A
Voltage across device V\(_R\) = 120v
Voltage across capacitor = V\(_C\) = √[V\(^2\) - V\(_R\)\(^2\)]
V\(_C\) = √[240\(^2\) - 120°]
V\(_C\) = 207.847v
I\(_C\) = I = 0.5A
X\(_C\) = V\(_C\) / I\(_C\)
= \(\frac{207.846}{0.5) → 415.7Ω
Capicitance, C = \(\frac{X_c}{2πf}\)
= \(\frac{415.7}{2 * 3.142 * 50}\)
= 1.323F
The capacitance C of a capacitor is the ratio of the amount of charge q on its plates to the potential difference between them, i.e C = q/v
(ii) I - charge on both capacitors are the same q1 = q2
Total capacitance, C = \(\frac{C_1 C_2}{C_1 + C_2}\)
C = \(\frac{C_2}{1 + C_1/C_2}\) = \(\frac{C_2}{1 + V_1 + C_2}\)
C = \(\frac{C_2 V_2}{V_1 + C_2}\) = \(\frac{C_2 V_2}{2}\)
C = 1/2 = \(C_2 V_2\)
II - voltage across V\(_1\) = \(\frac{C_2}{C_1 + C_2}\) * V
V\(_1\) = \(\frac{1}{C_1_{C_2} + 1}\) * V
but \(\frac{C_1}{C_2}\) = \(\frac{C_1}{C_2}\) = \(\frac{5}{2}\)
V\(_1\) = \(\frac{1}{5}{2} + 1}\) * 2
= \(\frac{2}{7}\) * 2 = \(\frac{4}{7}\) V
Voltage across C\(_2\) : V\(_2\) = 2 = \(\frac{4}{7}\) = \(\frac{10}{7}\) V