single 3 Ω resistor, because. This 3 Ω
resistor is in series with the top 3 Ω resistor, giving an equivalent resistance
in the top branch of 3 Ω + 3 Ω = 6 Ω. Finally, this 6 Ω resistor is in parallel
with the bottom 3 Ω resistor, giving an overall equivalent resistance of 2 Ω,
because.
- D If each of the identical bulbs has resistance R, then the current through each
bulb is ε/R. This is unchanged if the middle branch is taken out of the parallel
circuit. (What will change is the total amount of current provided by the
battery.)
- D The equation P = I^2 R gives
P = (0.5 A)^2 (100 Ω) = 25 W = 25 J/s
Therefore, in 20 s, the energy dissipated as heat is
E = Pt = (25 J/s)(20 s) = 500 J
- B Resistors in series always share the same current, so we can eliminate (D)
and (E). Now, using Ohm’s law, V = IR, we see that if I is constant, then V is
proportional to R. Since RA = 4RB, we know that VA = 4VB.
