SSE JE Electrical Engineering 29 Oct 2020 Shift 2 Papers

Question : 163 of 200

Marks: +1, -0

Find the current in each branch of the given network if the total current is 2.55 A.

Solution:

Here,

\V =10 \V

\R _{ \a }=2 \Ω

\R _{ \b }=8 \Ω

\R _{ \c }=5 \Ω

\R _{ \x }= \R _{ \a }+ \R _{ \b }=2+8=10 \Ω

\R _{ \y }= \R _{ \c }+ \R =(5+ \R ) \Ω

\R _{ \eq }= \R _{ x } {||} \R _{ \y }={{ \R_\x\R_\y }}/{{ \R_\x+\R_\y }}

={10(5+ R )}/{10+5+ R }={10(5+ \R )}/{15+ \R }

Since,

\V=\I \R_{{\eq }}

⇒10=2.55 ×{10(5+ \R )}/{15+ \R }

⇒{15+ \R }/{5+ \R }=2.55

⇒15+ \R =12.75+2.55 \R

⇒1.55 \R =2.25

⇒\R =1.45 \Ω

\R _{ \y }=5+ \R =5+1.45=6.45 \Ω

Now, through current division Rule

\I _{1}={ \R _{ \y }}/{ \R _{ \x }+ \R _{ \y }} ×\I

={6.45}/{10+6.45} × 2.55

={6.45 ×2.55}/{16.45}=1 \A

\I _{2}={ \R _{ \x }}{ \R _{ \x }+ \R _{ \y }} I

={10}/{10+6.45} 2.55

={10 ×2.55}/{16.45}=1.55 \A

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