Question

If the current in an electric bulb increases by 1 %, what will be the change in the power of a bulb ?[Assume that the resistance of the filament of a bulb remains constant.](A) increases by 1 %
(B) decreases by 1 %(C) increases by 2 %
(D) decreases by 2 %

Answer 1

Answer:

Option (C): The power will increase by 2%.

Explanation:

Assumptions:

I = Current through the bulb.

V = Potential difference across the bulb.

R = resistance of the bulb.

We know,

Power of bulb is given by

$P = VI$.

Using Ohm's law,

$V = IR$.

Therefore,

$P = (IR) I = I^2 R$.

Now, the current is increased by 1%.

Let the new current by $I'$, such that,

$I' =I + ( 1\%\ \text{of}\ I) =I + \dfrac{1}{100} I = 1.01\ I$.

And the resistance is same.

Therefore, the new power will be

$P' = I'^2R\\= (1.01\ I )^2R\\= 1.01^2\ I^2R\\= 1.01^2\ P\\=1.0201\ P.$

The % change in power is given by

$\Delta P = \dfrac{(P'-P)}{P}\times 100\% \\=\dfrac{(1.0201P - P)}{P}\times 100\% \\=(1.0201-1)\times 100\% = 2.01\%.$

Since, $P'>P$, therefore the power has increased.

Thus, the correct option is (C).

Answer 2

Dear Student,

◆ Answer -

(C) increases by 2 %

● Explaination -

# Given -

∆I/I = 1%

# Solution -

Power of the electric bulb is given by formula -

P = I²R

As resistance is constant, change in power of bulb is given by -

∆P/P % = 2 ∆I/I

Substitute, ∆I/I = 1% (given),

∆P/P % = 2×1

∆P/P % = +2 %

Therefore, power of the electric bulb will increase by 2 %.

Hope that was useful..