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SS3: PHYSICS - 2ND TERM

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SS3: PHYSICS – 2ND TERM Simple A.C Circuit Theory Questions – Simple A.C. Circuit
Lesson 6, Topic 11
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Theory Questions – Simple A.C. Circuit

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Topic Content:

Theory Questions & Answers – Simple A.C. Circuit

Theory Questions – Simple A.C. Circuit:

1. The Figure below 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.

Screenshot 2025 01 29 at 09.28.41

The table shows the square of the impedance, Z 

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Question 1

i. The Figure below 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, Z2, corresponding to each value of f2;

f2/Hz2 198.8 400.0 600.3 800.9 900.0
Z2/Ω2 249.6 400.0 552.3 702.3 800.0

Write down the equation for Z in terms of f2, R2, and L2.

Solution

⇒ ZRL =  \( \scriptsize \sqrt{R^2 \; + \; X_L^2} \)

but \( \scriptsize X_L = 2\pi f L \)

⇒ ZRL =  \( \scriptsize \sqrt{R^2 \; + \; \left(2\pi f L\right)^2} \)

⇒ ZRL =  \( \scriptsize \sqrt{R^2 \; + \; 4\pi^2 f^2 L^2} \)

 

(ii) Plot a graph of Z2 against f2 and use it to deduce:

(I) L
(II) R

Given ⇒ \(\scriptsize \pi^2 = 10 \)

Working:

 

slope, m  =  \(\frac{Z_2^2 \: – \: Z_1^2}{f_2^2 \: – \: f_1^2}\\ =\frac{700 \: – \: 200}{800 \: – \: 130}\\ = \frac{500}{670}\\ = \scriptsize 0.746 \)

⇒ ZRL =  \( \scriptsize \sqrt{R^2 \; + \; 4\pi^2 f^2 L^2} \)

∴  \( \scriptsize Z_{RL}^2 = R^2 \; + \; 4\pi^2 f^2 L^2 \)

∴  \( \scriptsize Z_{RL}^2 =  4\pi^2 L^2 f^2  \; + \; R^2\)

(I) L

Compare with:

⇒ \( \scriptsize y = mx \:+\: c\)

m = \( \scriptsize 4\pi^2 L^2 \)

⇒ \( \scriptsize L^2 = \normalsize \frac{m}{4\pi^2 } \)

but from the question, \(\scriptsize \pi^2 = 10 \)

∴  \( \scriptsize L^2 = \normalsize \frac{m}{4(10)} \)

⇒ \( \scriptsize L^2 = \normalsize \frac{m}{40} \)

⇒ \( \scriptsize L = \normalsize \sqrt{ \frac{m}{40}} \)

from the graph m = 0.746

∴  \( \scriptsize L = \normalsize \sqrt{ \frac{0.76}{40}} \\ = \scriptsize \sqrt{0.019} \\ = \scriptsize 0.138 \:H \: or \: 138 \:mH \)

 

(II) R = \( \scriptsize \sqrt{intercept} \)

R = \( \scriptsize \sqrt{100} \\ = \scriptsize  10 \: \Omega \)

Question 2

(i) Explain resonance frequency as applied in RLC series circuit.

Answer

It is the frequency of oscillation of an RLC series circuit when the capacitance reactance is equal to the inductive reactance

i.e. Xc = XL

The impedance is equal to the resistance. When current is maximum, impedance is minimum.

(ii) Sketch a diagram to illustrate the vibration of frequency, f, with the resistance, R, the capacitive reactance, Xc, and the inductive reactance, XL, in RLC series circuit.

Answer

 

(iii) Using the diagram drawn in (a)(ii), state whether the current in the circuit leads, lags or is in phase with the supply voltage when:

(a) f = f0

Answer

  • At resonance “f = f0“, the current is in phase with the supply voltage. The inductive reactance (XL) of an inductor exactly cancels out the capacitive reactance (XC) of a capacitor in the circuit, leaving only the resistance (R) to impede the current flow. This means that the circuit behaves purely resistive, allowing maximum current flow.

 

(b) f < f0

Answer

  • When “f < f0” (frequency (f) is less than the resonant frequency, f0), the current leads the supply voltage, meaning the current reaches its peak before the voltage does; indicating a predominantly capacitive behaviour in the circuit, meaning the capacitor’s reactance is larger than the inductor’s reactance (XC > XL), causing the current to rise before the voltage reaches its peak. 

 

(c) f > f0

  • When “f > f0” (frequency is greater than the resonant frequency, f0), in an RLC circuit, the current will “lag behind” the supply voltage, indicating a predominantly inductive behaviour in the circuit. This is because at higher frequencies, the inductive reactance (XL) becomes dominant (XL > XC), causing the current to lag behind the voltage.
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