Question

if the strength of the current decrease by 4% then the power will be​

Answer 1

(a) 4% as power is directly proportional to current

Answer 2

Answer:

The power of the bulb will decrease nearly by 8%.

Therefore the option (b) is correct.

Explanation:

Consider the initial current, $I_{i}=100A$

When current decreases by 4%,

then final current, $I_{f}=96A$

Initial power, $P_{i}=I_{i}^{2}R$

$P_{i}=(100A)^{2}R\\P_{i}=10000R$

Final power, $P_{f}=(I_{f})^{2}R$

$P_{f}=(96)^{2}R\\P_{f}=9216R$

Change in power ΔP$=P_{f}-P_{i}=9216R-10000R=-784R$

% change in power =(IΔPI/$P_{i}$)×100$=\frac{784R}{10000R}*100$ =7.84% ≈ 8%