Heat dissipation across a resistance R is 500J, if 1.5 A is passed through resistance for 20s what will be heat generation if 3A current is passed across the same resistance for 20 sec

Answer

To solve the problem of heat generation across a resistance when different currents are passed through it, we can use the formula for heat dissipation:

\[ H = I^2 R T \]

where:
- \( H \) is the heat generated,
- \( I \) is the current,
- \( R \) is the resistance,
- \( T \) is the time.

### Step-by-Step Solution:

1. **Identify the given values for the first scenario:**
- Heat dissipation \( H_1 = 500 \, J \)
- Current \( I_1 = 1.5 \, A \)
- Time \( T = 20 \, s \)

2. **Set up the equation for the first scenario:**
\[ H_1 = I_1^2 R T \]
Substituting the known values:
\[ 500 = (1.5)^2 R (20) \]

3. **Calculate \( R \):**
\[ 500 = 2.25 R (20) \]
\[ 500 = 45 R \]
\[ R = \frac{500}{45} \approx 11.11 \, \Omega \]

4. **Identify the values for the second scenario:**
- Current \( I_2 = 3 \, A \)

5. **Set up the equation for the second scenario:**
\[ H_2 = I_2^2 R T \]
Substituting the known values:
\[ H_2 = (3)^2 R (20) \]

6. **Substituting \( R \) from the first scenario:**
\[ H_2 = 9 \cdot \left(\frac{500}{45}\right) \cdot 20 \]

7. **Calculate \( H_2 \):**
\[ H_2 = 9 \cdot \frac{500 \cdot 20}{45} \]
\[ H_2 = 9 \cdot \frac{10000}{45} \]
\[ H_2 = 9 \cdot 222.22 \approx 2000 \, J \]

8. **Convert to kilojoules:**
\[ H_2 = 2 \, kJ \]

### Final Answer:
The heat generation when a current of 3 A is passed through the same resistance for 20 seconds is **2 kJ**.