An alternating voltage V=140 sin 50 t is applied to a resistor of resistance 10 `Omega`. This voltage produces `triangleH` heat in the resistor in time `trianglet`. To produce the same heat in the same time, rquired DC current is

Options

A. 14 A
B. About 20 A
C. about 10 A
D. None of these

Explanation

To solve the problem step by step, we will determine the required DC current that produces the same amount of heat in the resistor as the given alternating voltage does.

### Step 1: Identify the given parameters
- The alternating voltage is given as \( V = 140 \sin(50t) \).
- The resistance \( R = 10 \, \Omega \).

### Step 2: Calculate the RMS value of the alternating voltage
The RMS (Root Mean Square) value of an AC voltage is given by:
\[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \]
where \( V_0 \) is the maximum voltage. Here, \( V_0 = 140 \, V \).

Calculating \( V_{\text{rms}} \):
\[ V_{\text{rms}} = \frac{140}{\sqrt{2}} \approx 99.0 \, V \]

### Step 3: Calculate the power dissipated in the resistor
The power \( P \) dissipated in a resistor when an AC voltage is applied is given by:
\[ P = \frac{V_{\text{rms}}^2}{R} \]
Substituting the values:
\[ P = \frac{(99.0)^2}{10} = \frac{9801}{10} = 980.1 \, W \]

### Step 4: Relate heat produced to power
The heat produced \( \Delta H \) in time \( \Delta t \) is given by:
\[ \Delta H = P \cdot \Delta t \]
For the DC case, the power is given by:
\[ P_{\text{DC}} = I^2 R \]
where \( I \) is the DC current.

### Step 5: Set the powers equal to find the DC current
To produce the same heat \( \Delta H \) in the same time \( \Delta t \), we have:
\[ P_{\text{AC}} \cdot \Delta t = P_{\text{DC}} \cdot \Delta t \]
Thus, we can equate the powers:
\[ 980.1 = I^2 \cdot 10 \]
Solving for \( I^2 \):
\[ I^2 = \frac{980.1}{10} = 98.01 \]
Taking the square root:
\[ I = \sqrt{98.01} \approx 9.9 \, A \]

### Final Answer:
The required DC current is approximately \( 10 \, A \).

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