For the two resistors, R1 and R2, in parallel to glow with the same brightness, they must be identical (assign that value of resistance R) and the current through each must be the same too (assign it the value I). The question states that the power dissipated by R3 must be equal to the power dissipated by R, which we have now set as I^2 R. Since the current through R3 is twice the current I, we require that:
(2 I)^2 R3 = I^2 R, or R3 = (1/4) R after a bit of algebra.
The convenience of converting to high voltage for power distribution and back to lower voltage for safe use has led to the almost universal adoption of ac as the means for transporting electrical energy from power plants to the home. In addition, motors of simpler design are possible for ac. In particular, three-phase motors offer numerous advantages not possible with dc motors (increase starting torque, more uniform torque).
The voltage source of 120 V directly across the 20 ohm resistor causes 6.0 A of current to flow through it. Since a total current of 10 A is stated, it must be the case that 4.0 A of current flows through the unknown resistance. With 120 V directly across it too, R=V/I = (120 V)/(4.0 A) = 30 ohm.
To check this answer, evaluate the effective resistance of a 20 ohm and a 30
ohm resistor in parallel, then the resulting current flow:
1/R.eff = 1/(20 ohm) + 1/(30 ohm) = 3/(60 ohm) + 2/(60 ohm) = 5/(60 ohm). Thus R.eff = (60/5) ohm = 12 ohm10 A of current results from 120 V across 12 ohm!
The effective resistance of the 5 ohm and 10 ohm resistors connected in series is found from:
1/R.eff = 1/(5 ohm) + 1/(10 ohm) = 2/(10 ohm) + 1/(10 ohm) = 3/(10 ohm). Thus R.eff = (10/3) ohm = 3.3 ohmThe effective resistance of 3.3 ohm from the two resistors in parallel, will in turn be in series with the two internal resistances, 1 ohm each. Thus the overall effective resistance in the circuit is 3.3 ohm + 1.0 ohm + 1.0 ohm = 5.3 ohm. The effective battery is 3.0 V since the battery polarities are oppositely directed. The current through each battery is thus I=V/R = (3.0 V)/(5.3 ohm) = 0.56 A.
The voltage across each internal resistance is the same, V=I R = (0.56A)(1.0 ohm) = 0.56 V. So the terminal voltage of the "9V" battery is 9.0 V - 0.56V = 8.44 V; the terminal voltage of the "6V" battery is 6.0 V + 0.56V = 6.56 V, since the current is going the "wrong" way through the battery, i.e. the "6V" battery is being charged by the "9V" battery.
Note that the difference in voltage between the discharging battery and the charging battery is 1.88 V, the voltage across the parallel effective resistance of 3.3 ohm when 0.56 A is flowing. This serves as a check of our calculations.
Actually the typical 12 fl. oz. can has a circumference of 21 cm, making its diameter close to 7 cm. The height is more like 12 cm! The can as specified would have a volume of 1/4 fl. oz.! Also, since the resistivity quoted is close to that of Play-Doh, I must assume that we are considering a "pretend" can made of solid Play-Doh.
Anyways, using the dimensions stated in the question results in a
cross-sectional area of about 3 sq. cm. The cylinder's resistance
can be found from resistance = resistivity x length / area;
R = (10 ohm-cm)x(3 cm)/(3 sq cm) = 10 ohm.
The effective resistance of the 15 ohm and 25 ohm cans connected in parallel is found from:
1/R.eff = 1/(15 ohm) + 1/(25 ohm) = 5/(75 ohm) + 3/(75 ohm) = 8/(75 ohm). Thus R.eff = (75/8) ohm = 9.4 ohmThe effective resistance of 9.4 ohm from the two cans in parallel, will in turn be in series with the 10 ohm can and the battery. The voltage of the battery has not been specified, so a numerical answer in volts is not possible. However, I can state that 9.4/19.4 or 48% of the voltage applied by the battery will be found across the 25 ohm can.
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