(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]
Explanation
a) Differences between a d. c. generator and a moving coil galvanometer.
| D. C. generator | moving coil galvanometer |
| converts mechanical energy to electrical energy | converts electrical energy to mechanical energy. |
| uses split rings or commutator | uses hair springs. |
| rotation of coil is continuous | rotation of coil is incomplete |
| uses carbon brushes (as terminals) | uses jeweled bearings (as terminals) |
(b) Eddy currents are currents induced in a conductor when subjected to varying magnetic field.
Any valid additional information e.g.
- eddy current flows in a circular path or closed loops;
- eddy current generates heat;
- eddy current cannot flow through gaps or slots;
- The currents move in such a direction as to oppose the change producing them.
Devices in which eddy currents are applied
- pointers of sensitive electric meters
- sensitive mass balances
- brakes in large electric motors
- speedometers in automobiles
- detection of cracks in railway tracks
- detection of metals
(c) Principle on which a potentiometer is based
When a steady current is allowed to pass through a uniform wire, equal lengths of the wire will have equal potential differences.
OR
The p.d across a length of a wire is (directly) proportional to the length provided the wire has a uniform cross section.
(d)
Vector diagram showing
V(_L) with arrow
V(_C) with arrow
V(_R) with arrow
(i) V(^2) = V(_R^2) + (V(_L) – V(_C))(^2)
110(^2) = 80(^2) + (40 – V(_C))(^2)
(40 – V(_C)) = (sqrt{(110 + 80) (110 – 80)})
40 – V(_C) = + 75.5
V(_C) = 40 + 75.5
V(_C) = 115.5V
(ii) X(_c) (frac{V_c}{I} = frac{1}{2 pi f c})
C = (frac{1}{2 pi fV_c})
C (frac{2}{2 times 3.14 times 60 times 115.5} = 45.9muF)
(iii) (V_L = IX_L = I times 2 pi fL})
L = (frac{V_l}{I times 1 times 2 pifVc})
C = (frac{2}{2 times 3.142 times 60 times 2})
= 0.053H