A `750 Hz, 20 V` source is connected to as resistance of `100 Omega` an inductance of `0.1803 H` and a capacitance of `10 muF` all in sereis.Calculate the time in which the resistance (thermalcapacity `2J//.^(@)C`) wil get heated by `10^(@)C`.

Options

A. 245s
B. 365 s
C. 418 s
D. 348 s

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Calculate the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
where \( f = 750 \, \text{Hz} \).

Calculating ω:
\[ \omega = 2\pi \times 750 = 1500\pi \, \text{rad/s} \]

### Step 2: Calculate the inductive reactance (X_L) and capacitive reactance (X_C)
The inductive reactance (X_L) is given by:
\[ X_L = \omega L \]
where \( L = 0.1803 \, \text{H} \).

Calculating \( X_L \):
\[ X_L = 1500\pi \times 0.1803 \approx 847.26 \, \Omega \]

The capacitive reactance (X_C) is given by:
\[ X_C = \frac{1}{\omega C} \]
where \( C = 10 \times 10^{-6} \, \text{F} \).

Calculating \( X_C \):
\[ X_C = \frac{1}{1500\pi \times 10 \times 10^{-6}} \approx 21.22 \, \Omega \]

### Step 3: Calculate the impedance (Z)
The total impedance (Z) in an LCR circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
where \( R = 100 \, \Omega \).

Calculating Z:
\[ Z = \sqrt{100^2 + (847.26 - 21.22)^2} \]
\[ Z = \sqrt{10000 + (826.04)^2} \approx \sqrt{10000 + 683,000} \approx \sqrt{693000} \approx 834 \, \Omega \]

### Step 4: Calculate the RMS current (I_RMS)
Using Ohm's law for AC circuits:
\[ I_{RMS} = \frac{V_{RMS}}{Z} \]
where \( V_{RMS} = 20 \, \text{V} \).

Calculating \( I_{RMS} \):
\[ I_{RMS} = \frac{20}{834} \approx 0.024 \, \text{A} \]

### Step 5: Calculate the power (P)
The power in an AC circuit is given by:
\[ P = V_{RMS} \cdot I_{RMS} \cdot \cos \phi \]
where \( \cos \phi = \frac{R}{Z} \).

Calculating \( \cos \phi \):
\[ \cos \phi = \frac{100}{834} \approx 0.120 \]

Calculating the power:
\[ P = 20 \cdot 0.024 \cdot 0.120 \approx 0.0576 \, \text{W} \, (\text{or J/s}) \]

### Step 6: Calculate the time (T) for the resistance to heat by \( 10^\circ C \)
The energy required to raise the temperature of the resistance by \( 10^\circ C \) is given by:
\[ Q = S \cdot \Delta T \]
where \( S = 2 \, \text{J/}^\circ C \) and \( \Delta T = 10^\circ C \).

Calculating \( Q \):
\[ Q = 2 \cdot 10 = 20 \, \text{J} \]

Using the relationship between power, energy, and time:
\[ P = \frac{Q}{T} \implies T = \frac{Q}{P} \]

Calculating \( T \):
\[ T = \frac{20}{0.0576} \approx 348 \, \text{s} \]

### Final Answer
The time in which the resistance will get heated by \( 10^\circ C \) is approximately **348 seconds**.

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