ai. Define the electric potential at a point in an electric field.
ii. An uncharged body, A, was charged electrostatically by a test charge, B, using the method of induction and the method of contact. State two differences between the two methods.
b. An important precaution during an electricity experiment is to open the circuit when no readings are being taken. Give two reasons for the stated precaution.
ci. Fig. 11.0 is a circuit diagram in which a coil of inductance, L, and a resistor of resistance, R, are connected to a variable alternating source of frequency, f.
The table shows the square of the impedance, \(Z^2\); corresponding to each value of \(ƒ^2\).
| \(ƒ^2\)/ \(Hz^2\) |
198.80 |
400.00 |
600.30 |
800.90 |
900.00 |
| \(Z^2\)/ \(Ω^2\) |
249.60 |
400.00 |
550.30 |
702.30 |
800.90 |
Write down the equation for Z in terms of \(f^2\), \(R^2\), and \(L^2\).
ii. Plot a graph \(Z^2 against \(f^2\) of and use it to determine the values of:
i. L
ii. R
[\(π^2\) = 10]
ai. Electric potential at a point is defined as the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field.
ii. Process of Charging:
Induction: Involves redistributing charges within an uncharged body without direct contact with a charged object.
Contact: Requires direct physical contact with a charged object to transfer charges.
- Nature of Final Charge:
Induction: Results in a charge of opposite polarity to the charged object initially brought close.
Contact: Results in a charge of the same polarity as the charged object with which contact is made.
b. - Safety: Opening the circuit prevents current from flowing, which reduces the risk of electric shock. This is especially important when working with high voltages or currents.
- Conservation of battery power: If the circuit is left closed, the current will continue to flow even when no readings are being taken. This can drain the battery and shorten its lifespan.
- Accurate readings: If the circuit is left closed, the current flowing through the circuit can cause the components to heat up. This can affect the accuracy of the readings.
- Protection of equipment: If the circuit is left closed, a surge of current could flow through the circuit if there is a fault. This could damage the equipment or even cause a fire.
c. \(Z^2 = R^2 + X{_L}^2\)
\(Z^2 = R^2 + (2πf_L)^2
\(Z^2 = R^2 + 4π^2f^2L^2\)
\(Z^2 = 4π^2f^2L^2 + R^2\)
∴ \(Z = \sqrt{4π^2L^2f^2 + R^2}\)
ii. from the equation
\(Z^2 = 4π^2f^2L^2 + R^2\)
using y = mx + c = y = \(Z^2\), x = \(f^2\)
m = \(4π^2L^2\) = slope of the grape
m = \(\frac{ Z_2^2 - Z_1^2}{f_2^2 - f_1^2} = \frac{ 700 - 200 }{ 800 - 140 }\)
\(4π^2L^2 = \frac{ 500}{ 660}\)
\(L^2 = \frac{ 500}{ 660} \times \frac{1}{ 4π^2} =\frac{ 500}{ 660} \times \frac{1}{4 \times 10}\) ( since \(π^2\) = 10)
\(L^2 = 0.0189\)
∴ L = \(\sqrt{ 0.0189}\) = 0.138H = 138mH
ai. Electric potential at a point is defined as the work done in bringing a unit positive charge from infinity to that point against the electrical forces of the field.
ii. Process of Charging:
Induction: Involves redistributing charges within an uncharged body without direct contact with a charged object.
Contact: Requires direct physical contact with a charged object to transfer charges.
- Nature of Final Charge:
Induction: Results in a charge of opposite polarity to the charged object initially brought close.
Contact: Results in a charge of the same polarity as the charged object with which contact is made.
b. - Safety: Opening the circuit prevents current from flowing, which reduces the risk of electric shock. This is especially important when working with high voltages or currents.
- Conservation of battery power: If the circuit is left closed, the current will continue to flow even when no readings are being taken. This can drain the battery and shorten its lifespan.
- Accurate readings: If the circuit is left closed, the current flowing through the circuit can cause the components to heat up. This can affect the accuracy of the readings.
- Protection of equipment: If the circuit is left closed, a surge of current could flow through the circuit if there is a fault. This could damage the equipment or even cause a fire.
c. \(Z^2 = R^2 + X{_L}^2\)
\(Z^2 = R^2 + (2πf_L)^2
\(Z^2 = R^2 + 4π^2f^2L^2\)
\(Z^2 = 4π^2f^2L^2 + R^2\)
∴ \(Z = \sqrt{4π^2L^2f^2 + R^2}\)
ii. from the equation
\(Z^2 = 4π^2f^2L^2 + R^2\)
using y = mx + c = y = \(Z^2\), x = \(f^2\)
m = \(4π^2L^2\) = slope of the grape
m = \(\frac{ Z_2^2 - Z_1^2}{f_2^2 - f_1^2} = \frac{ 700 - 200 }{ 800 - 140 }\)
\(4π^2L^2 = \frac{ 500}{ 660}\)
\(L^2 = \frac{ 500}{ 660} \times \frac{1}{ 4π^2} =\frac{ 500}{ 660} \times \frac{1}{4 \times 10}\) ( since \(π^2\) = 10)
\(L^2 = 0.0189\)
∴ L = \(\sqrt{ 0.0189}\) = 0.138H = 138mH
