A 750 Hz, 20 V (rms) source is connected to a resistance of 100 ohm, an inductance of 0.1803 H and a

Q: Why don't the inductor and capacitor consume any power?

In ideal inductors and capacitors, the phase difference between voltage and current is exactly 90 degrees. Since average power is $V_{rms} I_{rms} \cos\phi$, and $\cos(90^\circ) = 0$, they consume zero average power. They only store and release energy.

Q: Why do we use $V_{rms}$ instead of peak voltage in the power formula?

AC power formulas naturally use RMS values because RMS (Root Mean Square) is defined as the equivalent DC value that would produce the same heating effect in a resistor. Using peak voltage would require an extra factor of $1/2$.

Q: What is the difference between heat capacity and specific heat capacity here?

Heat capacity ($S$) is the heat required to raise the temperature of the *entire* object by 1 degree, so $H = S \Delta\theta$. Specific heat capacity ($c$) is per unit mass, where $H = mc \Delta\theta$. The question gives the total heat capacity, so we don't need the mass.

Q: Why is the impedance formula a vector sum with a square root?

Because the voltages across the resistor, inductor, and capacitor are not in phase. They act like vectors (phasors) rotating at the same frequency. The resistor voltage is in phase with current, while inductor and capacitor voltages are 90 degrees ahead and behind, respectively.

Q: What would happen if there was heat loss to the surroundings?

If there was heat loss, the temperature wouldn't rise linearly. We would have to use Newton's Law of Cooling, setting up a differential equation where the net rate of heat gain equals the electrical power minus the heat lost to the environment.

Question

A $750	ext{ Hz}$, $20	ext{ V (rms)}$ source is connected to a resistance of $100\ \Omega$, an inductance of $0.1803	ext{ H}$ and a capacitance of $10\mu	ext{F}$ all in series combination. The time in which the resistance (heat capacity $2	ext{ J/}^\circ	ext{C}$) will get heated by $10^\circ	ext{C}$ is close to. (Assume no loss of heat to the surroundings)

Options:

(A) $418	ext{ s}$

(B) $245	ext{ s}$

(C) $365	ext{ s}$

(D) $348	ext{ s}$

Sigma Percentile · Accepted Answer

Imagine a series LCR circuit connected to an alternating current source. Our mission is to find out exactly how long it takes for the resistor to heat up by $10^\circ	ext{C}$. This problem beautifully bridges the gap between alternating current circuit analysis and classical calorimetry.

### Analyzing the Setup

Let's start by noting down all our given parameters. We have an AC source with a frequency $f = 750	ext{ Hz}$ and an RMS voltage $V_{rms} = 20	ext{ V}$. The circuit components are a resistor $R = 100\ \Omega$, an inductor $L = 0.1803	ext{ H}$, and a capacitor $C = 10\ \mu	ext{F}$.

The resistor has a heat capacity $S = 2	ext{ J/}^\circ	ext{C}$, and we need to find the time $t$ to achieve a temperature rise $\Delta	heta = 10^\circ	ext{C}$.

### The Master Equation for Impedance

To find the heat produced, we first need to know the power dissipated. And to find the power, we must calculate the total opposition to the current, which is the impedance, $Z$. The impedance depends on the inductive reactance, $X_L$, and the capacitive reactance, $X_C$.

Let's calculate them:
$$X_L = 2\pi f L = 2\pi(750)(0.1803) \approx 849.6\ \Omega$$
$$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi(750)(10 	imes 10^{-6})} \approx 21.2\ \Omega$$

Notice how much larger the inductive reactance is! The circuit is highly inductive. Now, let's bring these values back into our impedance formula:
$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$
$$Z = \sqrt{100^2 + (849.6 - 21.2)^2} \approx \sqrt{10000 + 686246} \approx 834\ \Omega$$

### Power Dissipation and Final Calculation

With the impedance known, we can find the average power dissipated. Remember, in an LCR circuit, power is only lost across the resistor. The formula simplifies beautifully:
$$P_{avg} = V_{rms} I_{rms} \cos\phi = V_{rms} \left(\frac{V_{rms}}{Z}ight) \left(\frac{R}{Z}ight) = \frac{V_{rms}^2 R}{Z^2}$$

Let's plug in our numbers:
$$P_{avg} = \frac{(20)^2 	imes 100}{(834)^2} \approx 0.0575\ 	ext{J/s}$$

This power is essentially the rate at which heat is generated. The total heat required to raise the resistor's temperature by $\Delta	heta$ is given by its heat capacity $S$ times $\Delta	heta$. So, power times time equals $S$ times $\Delta	heta$:
$$P_{avg} 	imes t = S \Delta	heta$$
$$0.0575 	imes t = 2 	imes 10$$

Solving for $t$, we get:
$$t = \frac{20}{0.0575} \approx 348	ext{ s}$$

**And that is our final answer!** It takes approximately $348$ seconds for the resistor to heat up by $10^\circ	ext{C}$. Always remember to connect the electrical power loss directly to the thermal energy gained when dealing with such interdisciplinary problems.

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$T h e W a y F or w a r d$

The Sigma Insight: AC Circuits and Power in AC Circuits

Analyzing the Setup

The Master Equation for Impedance

Power Dissipation and Final Calculation

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Circuit Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Impedance Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reactances

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Reactances

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Impedance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Average Power

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Power

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Time Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward