Animated Solution for Physics - Alternating Current: A $0.07\text{ H}$ inductor and a $12\Omega$ resistor are connected in series to a $220\text{ V}$, $50\text{ Hz}$ AC source. The approximate current in the circuit and the phase angle between current and source voltage are, respectively. [Take, $\pi$ as $\frac{22}{7}$]

Imagine you are standing in front of a massive power grid. The energy pulsing through those wires isn't just a steady stream; it's a dynamic, oscillating wave of alternating current. In this thrilling journey, we are going to dive deep into the heart of an $RL$ series circuit.

We are given a simple yet elegant setup: an inductor with an inductance of $L=0.07\text{ H}$ and a resistor with a resistance of $R=12\Omega$. These two components are connected in series to an AC source that provides a voltage of $V=220\text{ V}$ at a frequency of $f=50\text{ Hz}$.

Our mission? To uncover the exact current flowing through this circuit and to determine the phase angle between the current and the source voltage.

Before we can find the current, we need to understand how much opposition this circuit offers to the flow of AC. Unlike a simple DC circuit where only resistance matters, an AC circuit with an inductor introduces a new kind of opposition called inductive reactance.

The inductor doesn't just sit there; it actively fights the changing current. The strength of this fight depends on how fast the current is changing, which is dictated by the frequency of the AC source. The formula for inductive reactance is beautifully simple:

We know that the angular frequency $\omega$ is related to the linear frequency $f$ by the equation $\omega =2\pi f$. Therefore, we can rewrite our reactance formula as:

Let's bring in our known values. We have $f=50\text{ Hz}$ and $L=0.07\text{ H}$. The problem also gives us a handy approximation for $\pi$, asking us to use $\frac{22}{7}$.

Substituting these into our equation, we get:

Notice how perfectly these numbers are designed to cancel out! We can rewrite $0.07$ as $\frac{7}{100}$. Let's see the magic happen:

The $100$ in the numerator cancels with the $100$ in the denominator. The $7$ in the numerator cancels with the $7$ in the denominator. We are left with a pristine, whole number:

Now we have two forms of opposition in our circuit: the pure resistance $R=12\Omega$ and the inductive reactance $X_L=22\Omega$. But here is the catch—we cannot simply add them together like regular numbers.

Why? Because the voltage across the resistor and the voltage across the inductor are out of phase by exactly $90^{\circ }$. They exist in different dimensions of time. To find the total opposition, known as the impedance ($Z$), we must use vector addition.

Imagine a right-angled triangle. The base represents the resistance $R$, and the perpendicular height represents the inductive reactance $X_L$. The hypotenuse of this triangle is our total impedance $Z$. According to Pythagoras' theorem:

Let's plug in our values:

Squaring these numbers gives us:

Adding them together, we find:

Now, $\sqrt{628}$ isn't a perfect square. But let's think like a physicist. We know that $25^2=625$. Since $628$ is incredibly close to $625$, we can safely approximate our impedance:

With the total impedance in hand, finding the current is a breeze. We apply the AC version of Ohm's Law, which states that the current $I$ is equal to the total voltage $V$ divided by the total impedance $Z$:

Substituting our values:

Let's do the math. $220$ divided by $25$ gives us exactly:

We have found the current! But our mission isn't over yet. We still need to find the phase angle $\varphi$.

Let's return to our impedance triangle. The phase angle $\varphi$ is the angle between the total impedance $Z$ and the resistance $R$. Using basic trigonometry, the tangent of this angle is the ratio of the perpendicular to the base:

Plugging in our values for $X_L$ and $R$:

We can simplify this fraction by dividing the numerator and the denominator by $2$:

To isolate the angle $\varphi$, we take the inverse tangent of both sides:

We have successfully navigated the complexities of the $RL$ series circuit. We calculated the inductive reactance, used vector addition to find the total impedance, applied Ohm's Law to find the current, and used trigonometry to determine the phase angle.

Our final results are a current of $8.8\text{ A}$ and a phase angle of ${\tan }^{-1}\left(\frac{11}{6}\right)$. Looking at our options, this perfectly matches option (a).

This problem is a beautiful reminder of how algebra, geometry, and physics intertwine to describe the invisible forces that power our world. Keep practicing, keep visualizing, and never lose your curiosity!