Animated Solution for Physics - Electromagnetic Induction: Comprehension Passage

Imagine you are standing in front of a real-world electrical workbench. You have a metal filament lamp—which, for all practical purposes in alternating current (AC) circuits, acts as a pure resistor. You wire this lamp in series with a capacitor, and you power the entire combination using a standard $200\text{ V}$, $50\text{ Hz}$ AC supply.

The problem gives us some fascinating clues. The lamp is glowing, consuming exactly $500\text{ W}$ of active power, and if you were to take a voltmeter and measure the potential difference strictly across the lamp, it would read $100\text{ V}$. Our mission is to decode the hidden parameters of this circuit: the exact capacitance $C$ of the capacitor and the phase angle $\varphi$ between the total current and the supply voltage.

In a direct current (DC) circuit, voltages in series simply add up algebraically. But in the dynamic world of AC circuits, we must respect the phase differences. The voltage across a pure resistor, $V_R$, is perfectly in phase with the current. However, the voltage across a capacitor, $V_C$, lags the current by exactly $90^{\circ }$.

Because of this perpendicular relationship, the total supply voltage $V$ is the vector sum (or phasor sum) of $V_R$ and $V_C$. Mathematically, this forms a right-angled triangle where the total voltage is the hypotenuse:

We know the total RMS voltage $V=200\text{ V}$ and the resistor voltage $V_R=100\text{ V}$. Let's substitute these into our master equation:

Taking the square root, we find the voltage drop across the capacitor:

Now that we have the voltages across both individual components, we can easily find the phase angle $\varphi$. In our phasor diagram, the tangent of the phase angle is the ratio of the perpendicular component (the capacitor voltage) to the horizontal component (the resistor voltage).

Substitute the values we just discovered:

What angle gives a tangent of $\sqrt{3}$? It is exactly $60^{\circ }$.

This beautifully answers the second part of our problem! The current leads the total voltage by $60^{\circ }$.

To find the capacitance, we must first understand the flow of electrons—the current. The lamp is the only component in this circuit that consumes real, active power. Capacitors only store and release energy; they do not consume it.

The active power $P$ consumed by the resistor is the product of the voltage across it and the RMS current flowing through it:

We are given that the lamp consumes $500\text{ W}$ and has $100\text{ V}$ across it. Let's plug these in:

We now know that a steady RMS current of $5\text{ A}$ is marching through our series circuit.

With the current known, we can determine the opposition the capacitor offers to this AC current. This opposition is called capacitive reactance, denoted as $X_C$. According to Ohm's law for AC circuits:

Substitute our known values:

But what exactly is capacitive reactance? It is inversely tied to the physical capacitance $C$ and the angular frequency $\omega$ of the AC source:

We know the frequency $f=50\text{ Hz}$. Let's set up the grand final equation:

Rearranging to solve for $C$:

Here, the problem throws us a wonderful mathematical lifeline. We are told to assume $\pi \sqrt{3}\approx 5$. Let's use this elegant approximation:

To convert Farads into the much more practical microfarads ($\mu \text{F}$), we multiply by $10^6$:

And there we have it! Through a logical sequence of phasor analysis, power calculations, and reactance formulas, we have completely decoded the circuit. The capacitance is $100$, and the phase angle is $60$.