To solve the problem step by step, we will follow these calculations:
### Step 1: Calculate the heat required to raise the temperature
The heat required (Q) can be calculated using the formula:
\[ Q = \text{Thermal Capacity} \times \Delta T \]
where:
- Thermal Capacity = 100 J/°C
- Change in Temperature (\(\Delta T\)) = 10°C
Substituting the values:
\[ Q = 100 \, \text{J/°C} \times 10 \, \text{°C} = 1000 \, \text{J} \]
### Step 2: Calculate the impedance (Z) of the circuit
The total impedance (Z) in an RLC series circuit at resonance can be calculated using:
\[ Z = R \]
where:
- R = 20 Ω
### Step 3: Calculate the reactance of the inductor (X_L) and capacitor (X_C)
1. **Calculate the inductive reactance (X_L)**:
\[ X_L = \omega L \]
where:
- \(\omega = 2\pi f\) and \(f = 2000 \, \text{Hz}\)
- \(L = \frac{0.125}{\pi} \, \text{H}\)
First, calculate \(\omega\):
\[ \omega = 2 \pi \times 2000 = 4000 \pi \, \text{rad/s} \]
Now, calculate \(X_L\):
\[ X_L = 4000 \pi \times \frac{0.125}{\pi} = 500 \, \Omega \]
2. **Calculate the capacitive reactance (X_C)**:
\[ X_C = \frac{1}{\omega C} \]
where:
- \(C = \frac{500}{\pi} \, \text{nF} = \frac{500 \times 10^{-9}}{\pi} \, \text{F}\)
Now, calculate \(X_C\):
\[ X_C = \frac{1}{4000 \pi \times \frac{500 \times 10^{-9}}{\pi}} = \frac{1}{4000 \times 500 \times 10^{-9}} \]
\[ = \frac{1}{2 \times 10^{-6}} = 500 \, \Omega \]
### Step 4: Check for resonance
Since \(X_L = X_C\), the circuit is at resonance, and thus:
\[ Z = R = 20 \, \Omega \]
### Step 5: Calculate the power (P) consumed in the circuit
The power can be calculated using:
\[ P = \frac{V^2}{R} \]
where:
- \(V = 20 \, \text{V}\)
Substituting the values:
\[ P = \frac{20^2}{20} = \frac{400}{20} = 20 \, \text{W} \]
### Step 6: Calculate the time (t) required to heat the resistance
Using the formula:
\[ t = \frac{Q}{P} \]
Substituting the values:
\[ t = \frac{1000 \, \text{J}}{20 \, \text{W}} = 50 \, \text{s} \]
### Final Answer
The time in which the resistance will get heated by \(10°C\) is **50 seconds**.
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