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Find current through the circuit. Also, find heat generated in each resistor.


In Parallel,

1/Rp = 1/12 + 1/6 = (1+2)/12 = 4/12

Rp=3 ohm

Total Resistance = 4 + 4 ohms = 8 ohms

Current, I = V/R = 16/8 = 2A

Total Heat:- H= I2Rt

H = 22 x 8 x (1 second) = 4 x 8 = 32 J

I know for total heat, but how to find heat in individual resistor?


closed as off-topic by Hritik Narayan, HDE 226868, hft, Sebastian Riese, DanielSank Sep 20 '15 at 18:11

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  • 1
    $\begingroup$ Hi Harshan01. Welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems at meta.physics.stackexchange.com/questions/6093/… $\endgroup$ – SchrodingersCat Sep 20 '15 at 17:29
  • $\begingroup$ This is 100 % pure homework, at least show us some attempt at working out the problem. $\endgroup$ – Gert Sep 20 '15 at 17:29

The overall resistance of the circuit is 8 ohms. $$ I = \frac{V}{R} \\ I = \frac{16}{8} \\ I = 2 \textrm{ A}$$

2 A will flow through the first four ohm resistor, 1.33 A through the 12 ohm and 0.67 A through the 6 ohm resistor. You can now calculate the power dissipated in each resistor by using $P=I^2R$.

  • $\begingroup$ But Heat energy is H=I<sup>2</sup>Rt , right? $\endgroup$ – Harshan01 Sep 20 '15 at 17:42
  • $\begingroup$ @Harshan01 Yes. But then you'd need to know the amount of time that the circuit was left switched on for, and you didn't provide this. $\endgroup$ – Alby Sep 20 '15 at 17:44
  • $\begingroup$ Yes, I understand. 1 second of energy is equal to power, right? $\endgroup$ – Harshan01 Sep 20 '15 at 17:46
  • $\begingroup$ @Harshan01 "1 second of energy is equal to power" is totally meaningless. They are entirely different concepts. $\endgroup$ – Alby Sep 20 '15 at 17:48
  • $\begingroup$ I mean in this case, if you take time as 1 second, you still arrive at same results. $\endgroup$ – Harshan01 Sep 20 '15 at 17:50

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