(a) State two essential differences between a... | WAEC SSCE - Physics - 2009

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.

(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

(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 \(90^\circ\).

Phasor diagram showing the resistor voltage in phase with the current, the inductor voltage upward, the capacitor voltage downward, and the supply voltage as the resultant.

\[V^2=V_R^2+(V_L-V_C)^2\]

\[110^2=80^2+(40-V_C)^2\]

\[(40-V_C)^2=12100-6400=5700\]

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:

\[X_L=\frac{V_L}{I}=\frac{40}{2}=20\ \Omega\]

\[L=\frac{X_L}{2\pi f}=\frac{20}{2\times3.14\times60}=0.0531\text{ H}\]

\[\boxed{L\approx0.053\text{ H}}\]

Read lesson note on Scalars And Vectors (JAMB)

Answer Details