Question

When a potential difference across an electric bulb is $75V$, it draws a current of $3.5A$. What will be the change in current drawn by the bulb, if the potential difference is increased to $80V$?

• A. $2.333A$
• B. $0.233A$
• C. $233A$
• D. $0.0233A$

Answer 1

The correct option is B $0.233A$

Given:
Initial electric current drawn by the bulb $I_1=3.5A$
Initial potential difference, $V_1=75V$
New potential difference, $V_2=80V$

Let the resistance of the bulb be $R$ and the new current drawn be $I_2$ .

Using ohm's law,
$V_1=I_{1}R$ ......(1)
$V_2=I_{2}R$ ......(2)

On dividing equation $\left(2\right)$ by equation $\left(1\right)$, we get

$\frac{V_2}{V_1}=\frac{I_{2}R}{I_{1}R}$

$\Rightarrow \frac{V_2}{V_1}=\frac{I_2}{I_1}$

$\Rightarrow \frac{80}{75}=\frac{I_2}{3.5}$

$\Rightarrow I_2=3.73A$

$\therefore$ Increase in current,
$\Delta I=I_2-I_1$
$\Delta I=3.73-3.5$
$\Delta I=0.23A$