8.5.P23

In the circuit shown at the right, RA is identical to RB, and their resistance is half of RC: RA = RB = ½ RC. The current through resistor A is iA, iB is the current through resistor B, and iC through resistor C. The potential drop across resistor A is ΔVA and so on. The battery provides an EMF = ΔV0.

1) Which of the following must be true of the currents in the circuit?  

a) i_A = i_B, only

b) i_A = i_C, only

c) i_A = i_B = i_C

d) i_A = i_B + i_C

e) i_A = i_B – i_C

f) None of these.

2) What is the relationship between i_B and i_C?

a) i_B = ⅓ i_C

b) i_B = ½ i_C

c) i_B = i_C

d) i_B = 2i_C

e) i_B = 3i_C

f) None of these

3) Which of the following must be true about the potential drops in the circuit?

a) ΔV_A = ΔV_B

b) ΔV_B = ΔV_C

c) ΔV_A = ΔV_B

d) ΔV₀ = ΔV_B + ΔV_C

e) ΔV₀ = ΔV_A + ΔV_B

f) ΔV₀ = ΔV_A + ΔV_C

Joe Redish 4/19/13