NEET Questions Solved

(1) Applying Kirchoff's law in following figure.

At junction A :

$i+i_1+i_2=1$  .... (i)

For Loop (i)

$-60i+(15+5)i_1=0$

⇒ $i_1=3i$ ...(ii)

For loop (2)

– (15 + 5) i₁ + 10 i₂ = 0

⇒ i₂ = i₁ = (3 i) = 6i

On solving equation (i), (ii) and (iii) we get i = 0.1 A

Short Trick : Branch current = $\text{\hspace{0.17em}main current }\left(\frac{{\displaystyle \text{Resistance of opposite branch}}}{{\displaystyle \text{Total resistance}}}\right)$

⇒ $i=1\times \left[\frac{{\displaystyle \frac{{\displaystyle 20}}{{\displaystyle 3}}}}{{\displaystyle \frac{{\displaystyle 20}}{{\displaystyle 3}}+60}}\right]$ 

= 0.1 A