Question

for the circuit in Fig 10.3, V= 12V, R1= 4 ohms, R2=6ohms, and R3= 3 ohms. show that the power supplied by the battery (P=IV) is equal to that dissipated in the resistors (I^2R). what principle does this illustrate? Use the accompanying table. ( consider values significant to two decimal places.)

Answer 1

Total resistance of R2 and R3 in parallel is given by:
1/RT = 1/6 + 1/ 3 = 1/2
RT = 2ohms

The total circuit resistance (R1 and RT in series) = R1 + RT = 4 + 2 = 6ohms.

Current = V/R = 12/6 = 2A

Power supplied by battery = VI = 12x2 = 24W

Power dissipated by resistors = I^2R = 2^2 x 6 = 24W

This shows that power supplied = power dissipated.

Answer 2

R1 = 4?, R2 = 6?, R3 = 3?

?R_p= 6*3/9 = 2?

R = 4 + 2 = 6?

V = 12V

I = V/R = 2A

Power supplied by battery = VI = 24W

Power dissipated by R1 = 4*2² = 16W

Power dissipated by R2 = 3*(2*6/9)² = 16/3 W

Power dissipated by R1 = 6*(2*3/9)² = 8/3 W

Thus total power dissipated = 16 + 16/3 + 8/3 = 24 = power supplied by battery