A resistor dissipates 192 J of energy in 1 s when a current of 4 A is passed through it . Now , when the current is doubled ,the amount of thermal energy dissipated in 5 s is __________J.

Answer

To solve the problem step by step, we will follow the principles of electrical energy dissipation in a resistor.

### Step 1: Understand the given information
We know that a resistor dissipates 192 J of energy in 1 second when a current of 4 A is passed through it.

### Step 2: Use the formula for energy dissipation
The formula for the energy dissipated in a resistor is given by:
\[ E = I^2 R T \]
where:
- \(E\) is the energy in joules,
- \(I\) is the current in amperes,
- \(R\) is the resistance in ohms,
- \(T\) is the time in seconds.

### Step 3: Calculate the resistance
From the information provided, we can rearrange the formula to find the resistance \(R\):
\[ 192 = (4^2) R (1) \]
\[ 192 = 16R \]
\[ R = \frac{192}{16} = 12 \, \Omega \]

### Step 4: Calculate energy when current is doubled
Now, we need to find the energy dissipated when the current is doubled. The new current \(I' = 2 \times 4 = 8 \, A\).

### Step 5: Substitute the new values into the energy formula
Now we will calculate the energy dissipated when the current is 8 A for 5 seconds:
\[ E' = (I')^2 R T \]
Substituting the values:
\[ E' = (8^2) \times 12 \times 5 \]
\[ E' = 64 \times 12 \times 5 \]

### Step 6: Calculate the final energy
Now, we calculate:
\[ E' = 64 \times 12 = 768 \]
\[ E' = 768 \times 5 = 3840 \, J \]

### Final Answer
The amount of thermal energy dissipated in 5 seconds when the current is doubled is **3840 J**.

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