Compound Direct Current (DC) Circuits: Practice Questions (Physics 2 11.5)

Answer: (A). 1/R_eq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1 Ω⁻¹, so R_eq = 1.0 Ω. This is smaller than the smallest resistor, as it must be for a parallel group.

• (B) adds the resistances, which is the series rule.
• (C) is the mean of the three values, 11 Ω ÷ 3. There is no averaging rule.
• (D) is 1/11 Ω: it adds the resistances and then inverts, mixing the two rules.

Answer: (A). Each bulb is connected directly across the ideal battery, so each has the full emf across it before and after Z is added. The current in X, ΔV/R, does not change, so X is equally bright. The extra branch adds another equal current, so the current from the battery rises by a factor of 3/2.

• (B) treats the battery as supplying a fixed current that is shared out. An ideal battery fixes the potential difference, not the current.
• (C) thinks an extra resistor always adds resistance. In parallel it lowers R_eq.
• (D) is right about the battery current but wrong about X: its potential difference, and so its current, are unchanged. (With a real battery X would dim slightly, because the larger current would lower the terminal voltage.)

Answer: (A). ΔV_terminal = ℰ − Ir, so r = (ℰ − ΔV_terminal)/I = (9.0 V − 8.4 V) ÷ 0.30 A = 0.60 V ÷ 0.30 A = 2.0 Ω.

• (B) is 8.4 V ÷ 0.30 A: that is the external resistance, not the internal one.
• (C) is 9.0 V ÷ 0.30 A: the total resistance of the circuit, external plus internal.
• (D) multiplies the lost voltage by the current (0.60 V × 0.30 A). That gives a power in watts, not a resistance.

Answer: (A). Current is measured through the element, so the ammeter goes in series. Potential difference is measured between its ends, so the voltmeter goes in parallel. The real voltmeter adds a parallel branch, which lowers the resistance of that part of the circuit. That part then takes a smaller share of the potential difference, so the reading is slightly low.

• (B) has the right connections but the wrong direction of the effect. A parallel branch can only lower the resistance.
• (C) swaps the meters. An ammeter in parallel would short-circuit R, and a voltmeter in series would almost stop the current. Real meters always change the circuit a little.
• (D) puts the voltmeter in series. With its very large resistance it would carry almost no current, and it would not measure the potential difference across R.

1. Branch R₃ + R₄ = 4.0 Ω + 8.0 Ω = 12 Ω.
2. Parallel group: 1/R_p = 1/6.0 + 1/12 = 3/12, so R_p = 4.0 Ω.
3. (a) R_eq = 8.0 Ω + 4.0 Ω = 12 Ω.
4. (b) I₁ = 24 V ÷ 12 Ω = 2.0 A.
5. Potential difference across the parallel group: ΔV_p = (2.0 A)(4.0 Ω) = 8.0 V. (Check: ΔV₁ = 16 V, and 16 V + 8.0 V = 24 V.)
6. (c) I₂ = 8.0 V ÷ 6.0 Ω = 1.33 A.
7. Current in the R₃–R₄ branch: 8.0 V ÷ 12 Ω = 0.667 A. (Check: 1.33 A + 0.667 A = 2.0 A.)
8. (d) ΔV₃ = (0.667 A)(4.0 Ω) = 2.67 V.

Suggested mark points (4): 1 for R_eq = 12 Ω with the R₃ + R₄ branch combined first; 1 for I₁ = 2.0 A; 1 for I₂ = 1.33 A using ΔV_p = 8.0 V; 1 for ΔV₃ = 2.67 V. Accept ΔV₃ found by sharing 8.0 V between R₃ and R₄ in the ratio 4 : 8.

Common error: combining R₂ with R₃ alone in parallel. R₃ is not in parallel with R₂, because its ends are not both connected to the junctions; only the whole R₃–R₄ branch is.

(a) All of the current passes through A, and it then splits between the two branches. Both branches have the same potential difference, so the branch with the smaller resistance (B, resistance R) carries more current than the C–D branch (resistance 2R). C and D are in series, so they carry the same current. Ranking: A > B > C = D. (Values: I_A = 3ℰ/5R, I_B = 2ℰ/5R, I_C = I_D = ℰ/5R.)

(b) The wire short-circuits D, so D goes out. The C–D branch now has resistance R instead of 2R, so the parallel group falls from 2R/3 to R/2 and the total resistance falls from 5R/3 to 3R/2.

• A gets brighter: the total resistance is lower, so the current from the battery (all through A) increases.
• B gets dimmer: A now takes a larger share of the emf, so the potential difference across the parallel group, and the current in B, decrease (from 2ℰ/5R to ℰ/3R).
• C gets brighter: its branch resistance halved, so its current rises (from ℰ/5R to ℰ/3R), even though the group’s potential difference fell.

(c) R_eq = R + R/2 = 3R/2, so I_A = ℰ ÷ (3R/2) = 2ℰ/3R.

Accept correct reasoning in (b) that uses calculated currents throughout. Do not award the B point for “B dims because current is shared with C” without reference to the change in potential difference or calculated currents.

(a) I = ΔV/R: 5.0 V ÷ 5.0 Ω = 1.0 A; 4.4 V ÷ 2.0 Ω = 2.2 A.

(b) ΔV_terminal = ℰ − Ir for both: ℰ − (1.0 A)r = 5.0 V and ℰ − (2.2 A)r = 4.4 V. Subtract: (1.2 A)r = 0.60 V, so r = 0.50 Ω. Then ℰ = 5.0 V + (1.0 A)(0.50 Ω) = 5.5 V. Check: 5.5 V ÷ (2.0 Ω + 0.50 Ω) = 2.2 A. ✓

(c) A straight line sloping down. It crosses the voltage axis at ℰ = 5.5 V (zero current) and has slope −r = −0.50 V/A. (It would reach zero voltage at I = 11 A.)

(d) The claim is only partly right. The wire’s resistance is tiny, but the internal resistance is still in series with it. With the wire’s resistance negligible, I = ℰ/r = 5.5 V ÷ 0.50 Ω = 11 A. That is large and dangerous, but finite: the internal resistance limits it. (Wire resistance can be ignored here only in comparison with r, the one other resistance in the loop.)

Carry forward an error from (b) into (c) and (d) once.