Resistors 4Ω,6Ω and 12Ω are connected in parallel, its equivalent resistance (R) is given by R1=41+61+121⇒R=612=2Ω Again R is connected to 1.5 V battery whose internal resistance r=1Ω. Equivalent resistance now, R′=2Ω+1Ω=3Ω Current, Itotal =R′V=31.5=21 A Itotal =21=3x+2x+x=6x⇒x=121 ∴ Current through 4Ω resistor =3x =3×121=41 A Therefore, rate of Joule heating in the 4Ω resistor =I2R=(41)2×4=41=0.25 W
Three resistors of 4 Ω, 6 Ω and 12 Ω are connected in parallel and the combination is connected in series with a 1.5 ~V battery of 1 Ω internal…
JEE Main 2012 — Physics Electromagnetism
Three resistors of 4Ω,6Ω and 12Ω are connected in parallel and the combination is connected in series with a 1.5 V battery of 1Ω internal resistance. The rate of Joule heating in the 4Ω resistor is
Held on 12 May 2012 · Verified 6 Jul 2026.
Options
- A
0.55 W
- B
0.33 W
- C
0.25 W
- D
0.86 W
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