Question

In above shown figure, a circuit has resistor R and battery of emf $\epsilon$. If current I is flowing through circuit then power dissipated as heat is P. The circuit is changed by doubling the emf $\epsilon$ of the battery while R is kept constant. After the change, the current is:

• A. $\frac{I}{4}$
• B. $\frac{I}{2}$
• C. $I$
• D. $2I$
• E. $4I$

Answer 1

The correct option is D $2I$

Given - resistance in circuit $R_1=R$ , emf ${\epsilon }_1=\epsilon$ and current $I_1=I$

by applying ohm's law for given values

$\epsilon =IR$ ...........$.eq\left(1\right)$

when emf is doubled and resistance is kept constant

$R_2=R$ , ${\epsilon }_2=2\epsilon$ , $I_2=?$

$2\epsilon =I_{2}R$ ...............$eq\left(2\right)$

dividing $eq\left(2\right)$ by $eq\left(1\right)$

$I_2=2I$