Question

A light bulb and an open coil inductor are connected in series across an ac source of variable frequency. How will the glow of the bulb be affected when :
(I) an iron bar is inserted inside the coil, and
(II) the frequency of the source is decreased ?
Justify your answers. Assume that in each above case other factors remain unchanged.

Hint:

Remember the direct dependencies: - Core insertion $\rightarrow L \uparrow \rightarrow X_L \uparrow \rightarrow Z \uparrow \rightarrow I \downarrow \rightarrow \text{Glow } \downarrow$ - Frequency change $\rightarrow f \downarrow \rightarrow X_L \downarrow \rightarrow Z \downarrow \rightarrow I \uparrow \rightarrow \text{Glow } \uparrow$ Always trace your logic sequentially: Parameter $\rightarrow$ Reactance $\rightarrow$ Impedance $\rightarrow$ Current $\rightarrow$ Power/Glow.

Answer 1

Concept: In an alternating current (AC) circuit containing a resistance (such as a light bulb) and an inductive component (an open coil inductor) connected in series, the total opposition to the flow of current is determined by the impedance (\(Z\)) of the circuit. The circuit can be modeled as a series $RL$ circuit.
• Inductive Reactance (\(X_L\)): The opposition offered by an inductor to alternating current depends directly on the angular frequency (\(\omega\)) or linear frequency (\(f\)) of the source and the self-inductance (\(L\)) of the coil. It is mathematically given by: \[ X_L = \omega L = 2\pi f L \]
• Total Impedance (\(Z\)): For a series combination of a resistor \(R\) and an inductor \(L\), the net impedance is the vector sum of resistance and inductive reactance: \[ Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + (2\pi f L)^2} \]
• Alternating Current (\(I\)): The root-mean-square (rms) current flowing through the circuit is inversely proportional to the total impedance: \[ I = \frac{V}{Z} \]
• Brightness/Glow of the Bulb: The brightness or thermal power dissipated by the bulb is given by \(P = I^2 R\). Therefore, the glow of the bulb is directly proportional to the square of the current (\(I\)) flowing through it. Any factor that increases impedance decreases current and diminshes the glow, and vice versa. Step-by-step Justification for Case (I): An iron bar is inserted inside the coil.

• Let the initial self-inductance of the open air-core coil inductor be \(L_0\), given by the structural formula \(L_0 = \mu_0 n^2 A l\), where \(\mu_0\) is the permeability of free space, \(n\) is the number of turns per unit length, \(A\) is the cross-sectional area, and \(l\) is the length of the solenoid.
• When a ferromagnetic iron bar with a high relative magnetic permeability (\(\mu_r \gg 1\)) is inserted completely into the core of the inductor, the total magnetic permeability of the core material becomes \(\mu = \mu_r \mu_0\).
• Consequently, the new self-inductance \(L\) increases drastically, as defined by: \[ L = \mu_r L_0 \]
• Since the inductive reactance is directly proportional to the self-inductance (\(X_L = 2\pi f L\)), an increase in \(L\) directly leads to an increase in \(X_L\).
• Looking at the total impedance expression, \(Z = \sqrt{R^2 + X_L^2}\), an elevated value of \(X_L\) causes the net impedance \(Z\) of the series circuit to increase.
• From Ohm's law for AC circuits, the current is \(I = \frac{V}{Z}\). Since the source voltage \(V\) remains constant while the impedance \(Z\) increases, the rms current \(I\) traveling through the circuit must decrease.
• Because the power dissipation in the bulb is given by \(P = I^2 R\), a drop in current means less energy per second is converted into light. Thus, the glow of the bulb will decrease (diminish). Step-by-step Justification for Case (II): The frequency of the source is decreased.

• The inductive reactance of the coil is fundamentally defined by the linear relation with source frequency: \[ X_L = 2\pi f L \]
• If the frequency (\(f\)) of the variable alternating current source is deliberately decreased while holding the self-inductance \(L\) constant, the inductive reactance \(X_L\) will decrease linearly in proportion to the frequency drop.
• Now evaluating the total opposition, \(Z = \sqrt{R^2 + X_L^2}\), a reduction in the value of inductive reactance \(X_L\) causes a corresponding reduction in the overall circuit impedance \(Z\).
• According to the standard formulation for current, \(I = \frac{V}{Z}\), when the denominator impedance \(Z\) decreases under a steady voltage amplitude \(V\), the resulting root-mean-square alternating current \(I\) in the series circuit must increase.
• Since the electric current traversing the bulb filament rises, the electrical power converted into light and heat (\(P = I^2 R\)) increases significantly. Consequently, the glow of the bulb will increase (become brighter).