In above shown figure, determine the relationship between $I_{1}$ and $I_{2}$ .

I_{1} = \frac{1}{2} I_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I_{1} = \frac{2}{3} I_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I_{1} = I_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I_{1} = \frac{3}{2} I_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I_{1} = 2 I_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $I_{1} = 2 I_{2}$
Let the voltage of the battery be $V$
From figure, $V_{A B} = V_{C D} = V$
$∴$ Current $I_{1} = \frac{V_{A B}}{1} = V$

Also $I_{2} = \frac{V_{C D}}{2} = \frac{V}{2}$ $⟹$ $I_{1} = 2 I_{2}$

Suggest Corrections

Similar questions

Now what should be the correct relation between $I_{1}$ , $I_{2}$ and $I_{3}$ .

f : R \to R

f (x) + f (x + 1) = 2,

x \in R .

I_{1} = \int_{0}^{8} f (x) d x

I_{2} = \int_{- 1}^{3} f (x) d x

I_{1} + 2 I_{2}

I_{1}

I_{2}

x

(2 - i) (1 - 2 i)^{2}

If $I_{1} = \int_{0}^{\frac{π}{2}} x s i n x d x & I_{2} = \int_{0}^{\frac{π}{2}} x c o s x d x$ , then which of the following is true?

Join BYJU'S Learning Program