(a) State two essential differences between a moving coil galvanometer and a d.c. generator. (b) Explain the term eddy currents and state two devices in whi...
Question 1 Report
(a) State two essential differences between a moving coil galvanometer and a d.c. generator.
(b) Explain the term eddy currents and state two devices in which the currents are applied.
(c) State the principle on which the potentiometer is based when it is functioning.
(d) A source of e.m.f. 110 V and frequency 60Hz is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 2A, the potential differences across the resistor is 80 V and that across the inductor is 40 V. Draw the vector diagram of the potential differences across the inductor, the capacitor and the resistor.
Calculate the:
(i) potential difference across the capacitor;
(ii) capacitance of the capacitor;
(iii) inductance of the inductor. [π = 3.14]
Answer Details
(a) Differences between a moving-coil galvanometer and a d.c. generator
| Moving-coil galvanometer | D.C. generator |
|---|---|
| It detects or measures small electric currents. | It generates electrical energy. |
| It converts electrical energy to mechanical deflection; its coil turns through a limited angle and is controlled by hair springs. | It converts mechanical energy to electrical energy; its coil rotates continuously and uses a split-ring commutator and carbon brushes. |
(b) Eddy currents
Eddy currents are circulating currents induced in the body of a conductor when the magnetic flux through it changes. They flow in closed paths and oppose the change producing them. They are applied in an induction furnace and an eddy-current brake.
(c) Principle of a potentiometer
When a steady current flows through a uniform wire of constant cross-sectional area, the potential difference across the wire is directly proportional to its length:
\[V \propto l.\]
(d) Vector diagram
Taking the current, and hence \(V_R\), as the horizontal reference, \(V_L\) leads the current by \(90^\circ\), while \(V_C\) lags the current by \(90^\circ\).
\[V^2=V_R^2+(V_L-V_C)^2\]
\[110^2=80^2+(40-V_C)^2\]
\[(40-V_C)^2=12100-6400=5700\]
Since the circuit is net capacitive, \(V_C>V_L\):
\[V_C=40+\sqrt{5700}=40+75.5=115.5\text{ V}\]
(i) Potential difference across the capacitor:
\[\boxed{V_C=115.5\text{ V}}\]
(ii) Capacitive reactance:
\[X_C=\frac{V_C}{I}=\frac{115.5}{2}=57.75\ \Omega\]
\[C=\frac{1}{2\pi fX_C}=\frac{1}{2\times3.14\times60\times57.75}=4.59\times10^{-5}\text{ F}\]
\[\boxed{C\approx46\ \mu\text{F}}\]
(iii) Inductive reactance:
\[X_L=\frac{V_L}{I}=\frac{40}{2}=20\ \Omega\]
\[L=\frac{X_L}{2\pi f}=\frac{20}{2\times3.14\times60}=0.0531\text{ H}\]
\[\boxed{L\approx0.053\text{ H}}\]
Download The App On Google Playstore
Everything you need to excel in your exams
