A circuit with an electrical load having impedance Z is connected with an AC source as shown in the

Q: Why do we use $400$ as $\omega$?

Because the standard form of AC voltage is $V(t) = V_m \sin(\omega t)$. Comparing this with $300 \sin(400t)$, we directly get $\omega = 400 \text{ rad/s}$.

Q: How do we know if the current leads or lags?

If $X_L > X_C$, the circuit is inductive, and current lags the voltage. If $X_C > X_L$, it is capacitive, and current leads.

Q: Why does Graph 5 correspond to a lagging current?

A lagging current has the form $\sin(\omega t - \phi)$. At $t=0$, this value is negative. Graph 5 starts at a negative value, confirming the lag.

Q: What happens at resonance?

At resonance, $X_L = X_C$. They cancel each other out, making the circuit purely resistive. The impedance is minimum ($Z=R$), and the current is maximum and in phase with the voltage.

Q: Why is the phase angle exactly $53^\circ$?

The reactances and resistance form a $30-40-50$ impedance triangle, which is a multiple of the classic $3-4-5$ right triangle. The angle opposite the side of length 4 is approximately $53.1^\circ$.

Question

A circuit with an electrical load having impedance $Z$ is connected with an AC source as shown in the diagram. The source voltage varies in time as $V(t) = 300 \sin(400t)$ V, where $t$ is time in s. List-I shows various options for the load. The possible currents $i(t)$ in the circuit as a function of time are given in List-II. Choose the option that describes the correct match between the entries in List-I to those in List-II.

List-I:

P. $R = 30 \,\Omega$

Q. $R = 30 \,\Omega$, $L = 100 	ext{ mH}$

R. $C = 50 \,\mu	ext{F}$, $R = 30 \,\Omega$, $L = 25 	ext{ mH}$

S. $C = 50 \,\mu	ext{F}$, $R = 60 \,\Omega$, $L = 125 	ext{ mH}$

List-II:

1. Graph 1

2. Graph 2

3. Graph 3

4. Graph 4

5. Graph 5

Sigma Percentile · Accepted Answer

The beauty of alternating current circuits lies in the interplay between resistance, inductance, and capacitance. In this problem, we are tasked with matching four different electrical loads to their corresponding current-time graphs. Let's embark on this journey by decoding the source and analyzing each load step-by-step.

### Decoding the Source

Our journey begins with the voltage source. The problem provides the voltage equation:
$$V(t) = 300 \sin(400t)$$

From this standard form $V(t) = V_m \sin(\omega t)$, we can immediately extract two vital parameters:
1. The peak voltage, $V_m = 300 	ext{ V}$.
2. The angular frequency, $\omega = 400 	ext{ rad/s}$.

This angular frequency is the heartbeat of our circuit. It dictates how the inductors and capacitors will react to the alternating current.

### Case P: The Pure Resistor

Let's examine the first load, Case P. It consists solely of a resistor with $R = 30 \,\Omega$.

In a purely resistive circuit, life is simple. There are no phase shifts to worry about. The voltage and current dance perfectly in sync. The total impedance $Z$ is just the resistance itself:
$$Z = R = 30 \,\Omega$$

Using Ohm's law, we can find the current:
$$i(t) = \frac{V(t)}{Z} = \frac{300}{30} \sin(400t) = 10 \sin(400t)$$

This equation tells us that the current is a sine wave starting at zero with a peak amplitude of $10 	ext{ A}$. Looking at our options, **Graph 3** perfectly matches this description. Thus, P matches with 3.

### Case Q: The Inductive Lag

Moving on to Case Q, we introduce an inductor into the mix. We have a resistor $R = 30 \,\Omega$ and an inductor $L = 100 	ext{ mH}$.

First, we must calculate the inductive reactance, which is the opposition offered by the inductor:
$$X_L = \omega L = 400 	imes 100 	imes 10^{-3} = 40 \,\Omega$$

Now, we construct our impedance triangle. With $R = 30$ and $X_L = 40$, we have a classic $3-4-5$ right triangle. The total impedance $Z$ is:
$$Z = \sqrt{R^2 + X_L^2} = \sqrt{30^2 + 40^2} = 50 \,\Omega$$

The phase angle $\phi$ is given by:
$$	an \phi = \frac{X_L}{R} = \frac{40}{30} \implies \phi = 53^\circ$$

Because this is an inductive circuit, the current **lags** the voltage. The current equation becomes:
$$i(t) = \frac{300}{50} \sin(400t - 53^\circ) = 6 \sin(400t - 53^\circ)$$

At $t=0$, the current is negative. **Graph 5** shows a sine wave with an amplitude of $6 	ext{ A}$ starting at a negative value, confirming our result. Thus, Q matches with 5.

### Case R: The Capacitive Lead

Case R presents a full RLC circuit: $C = 50 \,\mu	ext{F}$, $R = 30 \,\Omega$, and $L = 25 	ext{ mH}$.

Let's calculate both reactances to see who wins the tug-of-war:
$$X_C = \frac{1}{\omega C} = \frac{1}{400 	imes 50 	imes 10^{-6}} = 50 \,\Omega$$
$$X_L = \omega L = 400 	imes 25 	imes 10^{-3} = 10 \,\Omega$$

The net reactance is $X_{net} = X_L - X_C = 10 - 50 = -40 \,\Omega$. The negative sign indicates that the capacitor dominates!

The total impedance $Z$ is again:
$$Z = \sqrt{30^2 + (-40)^2} = 50 \,\Omega$$

Since the circuit is net capacitive, the current **leads** the voltage by $53^\circ$. The current equation is:
$$i(t) = \frac{300}{50} \sin(400t + 53^\circ) = 6 \sin(400t + 53^\circ)$$

At $t=0$, the current is positive. **Graph 2** depicts a wave with an amplitude of $6 	ext{ A}$ starting at a positive value. Thus, R matches with 2.

### Case S: The Magic of Resonance

Finally, we arrive at Case S: $C = 50 \,\mu	ext{F}$, $R = 60 \,\Omega$, and $L = 125 	ext{ mH}$.

Let's evaluate the reactances:
$$X_C = 50 \,\Omega \quad 	ext{(same as before)}$$
$$X_L = 400 	imes 125 	imes 10^{-3} = 50 \,\Omega$$

Look closely! $X_L = X_C$. The inductive and capacitive reactances are perfectly balanced and cancel each other out. The circuit is in a state of **electrical resonance**.

At resonance, the impedance is purely resistive and at its minimum:
$$Z = R = 60 \,\Omega$$

The current is perfectly in phase with the voltage:
$$i(t) = \frac{300}{60} \sin(400t) = 5 \sin(400t)$$

This is a pure sine wave with an amplitude of $5 	ext{ A}$. **Graph 1** is the exact visual representation of this equation. Thus, S matches with 1.

### The Final Verdict

By systematically calculating the impedance and phase angle for each load, we have successfully decoded the entire matrix:
- **P** $ightarrow$ **3**
- **Q** $ightarrow$ **5**
- **R** $ightarrow$ **2**
- **S** $ightarrow$ **1**

This problem beautifully illustrates how different components shape the flow of alternating current, creating a unique signature for every circuit configuration.

List-I

List-II

Select Matching Pairs:

Visualized Solution

\text{Source Parameters}

\text{Case P: Pure Resistor}

\text{Current in Case P}

\text{Case Q: RL Circuit}

\text{Impedance \& Current in Case Q}

\text{Case R: RLC Circuit}

\text{Impedance \& Current in Case R}

\text{Case S: Resonance}

\text{Current in Case S}

\text{Final Matching}

The Sigma Insight: Alternating Current (AC) and Voltage

Decoding the Source

Case P

Case Q

Case R

Case S

The Final Verdict

Similar Questions

List-I

List-II

In the given circuit, the AC source has $ω = 100 rad/s$ . Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

In $L - C$ circuit, the inductance $L = 40$ mH and capacitance $C = 100 μ F$ . If a voltage $V (t) = 10 sin (314 t)$ is applied to the circuit, the current in the circuit is given as

List-I

List-II

In the circuit shown, $L = 1 μ H$ , $C = 1 μ F$ and $R = 1 k Ω$ . They are connected in series with an AC source $V = V_{0} sin ω t$ as shown. Which of the following options is/are correct?

List-I

List-II

An alternating voltage $V (t) = 220 sin 100 π t$ volt is applied to a purely resistive load of $50Ω$ . The time taken for the current to rise from half of the peak value to the peak value is

An AC voltage source of variable angular frequency $ω$ and fixed amplitude $V_{0}$ is connected in series with a capacitance $C$ and an electric bulb of resistance $R$ (inductance zero). When $ω$ is increased,

List-I

List-II

Select Matching Pairs:

Visualized Solution

\text{Source Parameters}

\text{Case P: Pure Resistor}

\text{Current in Case P}

\text{Case Q: RL Circuit}

\text{Impedance \& Current in Case Q}

\text{Case R: RLC Circuit}

\text{Impedance \& Current in Case R}

\text{Case S: Resonance}

\text{Current in Case S}

\text{Final Matching}

The Sigma Insight: Alternating Current (AC) and Voltage

Decoding the Source

Case P

Case Q

Case R

Case S

The Final Verdict

Similar Questions

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the given circuit, the AC source has ω=100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In L−C circuit, the inductance L=40 mH and capacitance C=100μF. If a voltage V(t)=10sin(314t) is applied to the circuit, the current in the circuit is given as

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the circuit shown, L=1μH, C=1μF and R=1kΩ. They are connected in series with an AC source V=V0​sinωt as shown. Which of the following options is/are correct?

List-I

List-II

In the given circuit, the AC source has $ω = 100 rad/s$ . Considering the inductor and capacitor to be ideal, the correct choice(s) is(are)

In $L - C$ circuit, the inductance $L = 40$ mH and capacitance $C = 100 μ F$ . If a voltage $V (t) = 10 sin (314 t)$ is applied to the circuit, the current in the circuit is given as

In the circuit shown, $L = 1 μ H$ , $C = 1 μ F$ and $R = 1 k Ω$ . They are connected in series with an AC source $V = V_{0} sin ω t$ as shown. Which of the following options is/are correct?