The angular frequency of alternating current in an L-C-R circuit is 100 rad/s. The components connec

Q: Why did we use algebraic subtraction for KCL in an AC circuit?

You caught the trap! In reality, AC currents must be added as phasors because they have different phase angles. However, the official JEE solution for this specific question used simple algebraic subtraction. We followed it to get the intended answer, but always remember that physically, it's a flawed approach.

Q: How do we know $R'$ and $L$ are in parallel?

Look at the nodes. The current from the main branch splits at the top junction, with one path going through $R'$ and the other through $L$. Both paths then reconnect at the bottom wire. This is the classic definition of a parallel combination.

Q: Why is the voltage across $R'$ also 20 V?

Components in parallel share the exact same potential difference. Since the problem states the voltage across the inductor $L$ is 20 V, the voltage across the parallel resistor $R'$ must also be 20 V.

Q: Could we have found $C$ without finding $I$ first?

No, because the capacitance depends on the capacitive reactance ($X_C$), which requires both the voltage across the capacitor and the current flowing through it. Since $C$ is in series with $R$, they share the same current $I$.

Q: What does the negative sign in $I_2 = -1/4$ A mean?

In the context of the algebraic KCL used here, the negative sign simply implies that the assumed direction of current was opposite to the calculated one. We take the magnitude ($1/4$ A) to find the reactance, as reactance is a positive scalar quantity.

Question

The angular frequency of alternating current in an L-C-R circuit is $100	ext{ rad/s}$. The components connected are shown in the figure. Find the value of inductance of the coil and capacity of condenser.

Options:

(A) 0.8 H and $150\mu	ext{F}$

(B) 0.8 H and $250\mu	ext{F}$

(C) 1.33 H and $250\mu	ext{F}$

(D) 1.33 H and $150\mu	ext{F}$

Sigma Percentile · Accepted Answer

### Analyzing the Circuit Topology

Welcome to this fascinating AC circuit problem! Before diving into the math, we must carefully map out the circuit's topology. We have an AC source connected to a main branch that contains a capacitor $C$ and a resistor $R$ in series. After passing through these components, the circuit splits into two parallel branches: one containing a resistor $R'$ and the other containing an inductor $L$.

Understanding this structure is crucial because it dictates how current flows and how voltage is distributed across the components.

### Finding the Main Branch Current

To find the unknown capacitance and inductance, our first objective is to determine the currents flowing through the various branches. Let's start with the top resistor $R$. We are given its resistance as $60\ \Omega$ and the voltage across it as $15	ext{ V}$.

Using Ohm's law, we can easily find the main current $I$ flowing through it:

$$I = \frac{V_R}{R} = \frac{15}{60} = \frac{1}{4}	ext{ A}$$

This is the total current flowing through the main branch, which means this exact same current of $0.25	ext{ A}$ also passes through the capacitor $C$.

### Unlocking the Capacitance

Now, let's focus our attention on the capacitor. The voltage across it is given as $10	ext{ V}$, and we just established that the current flowing through it is $\frac{1}{4}	ext{ A}$.

The capacitive reactance $X_C$ is simply the ratio of voltage to current:

$$X_C = \frac{V_C}{I} = \frac{10}{1/4} = 40\ \Omega$$

Since we know that capacitive reactance is inversely proportional to the angular frequency and capacitance ($X_C = \frac{1}{\omega C}$), and we are given $\omega = 100	ext{ rad/s}$, we can calculate the capacitance $C$:

$$C = \frac{1}{\omega X_C} = \frac{1}{100 	imes 40} = \frac{1}{4000}	ext{ F} = 250\ \mu	ext{F}$$

### The Parallel Branch and the KCL Trap

Next, let's examine the parallel section of the circuit. The voltage across the inductor is given as $20	ext{ V}$. Because the resistor $R'$ is in parallel with the inductor, the voltage across $R'$ must also be exactly $20	ext{ V}$.

We can find the current $I_1$ flowing through $R'$:

$$I_1 = \frac{V_{R'}}{R'} = \frac{20}{40} = \frac{1}{2}	ext{ A}$$

**Here is where this specific JEE question contains a famous conceptual trap!** In a real AC circuit, currents in parallel branches containing different types of components (like a resistor and an inductor) have different phase angles. Therefore, they must be added as phasors.

However, the official solution for this problem uses simple algebraic subtraction, assuming direct scalar Kirchhoff's Current Law (KCL). To arrive at the intended answer, we must follow their logic. The current $I_2$ through the inductor is calculated as the total current $I$ minus $I_1$:

$$I_2 = I - I_1 = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}	ext{ A}$$

Taking the magnitude, the current through the inductor is $|I_2| = \frac{1}{4}	ext{ A}$.

### Calculating the Inductance

Finally, we can determine the inductance. The inductive reactance $X_L$ is the voltage across the inductor divided by the current flowing through it:

$$X_L = \frac{V_L}{|I_2|} = \frac{20}{1/4} = 80\ \Omega$$

Since inductive reactance is given by $X_L = \omega L$, we can solve for $L$:

$$L = \frac{X_L}{\omega} = \frac{80}{100} = 0.8	ext{ H}$$

### Final Conclusion

We have successfully determined both unknown values. The inductance is **$0.8	ext{ H}$** and the capacitance is **$250\ \mu	ext{F}$**. This perfectly matches option (b).

While the algebraic KCL step is physically flawed for AC circuits, recognizing how the examiner intended the problem to be solved is a vital skill for competitive exams. Always trust your fundamental concepts, but be prepared to navigate such anomalies!

Animated Solution for Physics - Alternating Current: The angular frequency of alternating current in an L-C-R circuit is $100 rad/s$ . The components connected are shown in the figure. Find the value of inductance of the coil and capacity of condenser.

Select Answer:

Visualized Solution

Circuit Topology

Main Branch Current

Calculate Main Current

Capacitance Calculation

Parallel Branch Current

The KCL Trap

Calculate Inductor Current

Inductance Calculation

Final Answer

The Sigma Insight: Alternating Current (AC) and Voltage

Analyzing the Circuit Topology

Finding the Main Branch Current

Unlocking the Capacitance

The Parallel Branch and the KCL Trap

Calculating the Inductance

Final Conclusion

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