Practicing Success
Practicing Success
CUET Preparation Today
The ratio of powers dissipated respectively in R and 3R, as shown is |
9 27/4 4/9 4/27 |
4/27 |
$\text{Current through resistance R is } \frac {2I}{3}$ $ \text{Power dissipated through R is } P_1 = \frac{4I^2}{9} \times R = \frac{4I^2R}{9}$ $ \text{Power dissipated through 3R is } P_2 = I^2.3R$ $ \frac{P_1}{P_2} = \frac{4}{27}$ |
