SSE JE Electrical Engineering 28 Oct 2020 Shift 1 Papers

Question : 199 of 200

Marks: +1, -0

Find

\R_\eq

for the following circuit.

Solution:

R _{1}=6 \Ω

R _{2}=2 \Ω

R _{3}=6 \Ω

R _{4}=2 \Ω

R _{34}= R _{3}+ R _{4}=6+2=8 \Ω

R '= R _{2} ∥R _{23}

=(2 ×8) /(2+8)=16 / 10=1.6 \Ω

R _{ eq }= R _{1}+ R '

=6+1.6=7.6 \Ω

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