1Ω and 3Ω are in series so, Rs=R1+R2Rs=3+1=4Ω and R2=8Ω net i= current in the circuit Current through R1 is i1=R1+R2i×R2 =12i×8=321 Current R2 is i2=R1+R21×R1=12i×4 =12i×8 Power dissipated in 3Ω resistance is: P1=l12×3 Power dissipated in 8Ω resistance is: P2=l22×8 For (i) and (ii) P2P1=i22×8i12×3⇒P2P1 ∴P1=(3i)2×8(32i)2×3=23=23×P2=23×2=3 W ∵∴⇒PPI=100 W,V=125V=VI=VP=125100A
Power dissipated across the 8 Ω resistor in the circuit shown here is 2 watt. The power dissipated in watt units across the 3 Ω resistor is: NEET…
NEET UG 2006 — Physics Electromagnetism
Power dissipated across the 8Ω resistor in the circuit shown here is 2 watt. The power dissipated in watt units across the 3Ω resistor is: 
Held on 30 Apr 2006 · Verified 9 Jul 2026.
Options
- A
3.0
- B
2.0
- C
1.0
- D
0.5
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