Since the network is balanced, wheatstone's bridge. Therefore no current flows through the arm AC. 
Here R1 and R2 are in series ∴R1′=R1+R2=4+2=6ΩR3 and R4 are in series R∗=R3+R4=6+3 =9Ω Now R′ and R′′ are in parallel ∴Req1=R′1+Rn1=61+91=185⇒Req=516Now from V=Ireq⇒I=ReqV=185 V
NEET UG 2005 — Physics Electromagnetism
2005mcqmedium
For the network shown in the figure the value of the current i is: 
Official previous-year question
Held on 30 Apr 2005 · Verified 9 Jul 2026.
Options
- A
359 V
- B
518 V
- C
95V
- D
185V
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