20 questions

I = Q / t = 15 ÷ 5 = 3 A

Convert to seconds first: 3 minutes = 180 s.

Q = I × t = 2 × 180 = 360 C

V = W / Q = 45 ÷ 9 = 5 V

Rearrange V = W/Q to make W the subject: W = V × Q.

W = 9 × 8 = 72 J

V = I × R = 2.5 × 8 = 20 V

R = V / I = 15 ÷ 3 = 5 Ω

(a) R_total = 4 + 6 = 10 Ω

(b) I = V / R = 20 ÷ 10 = 2 A — this is the current everywhere in the loop

(c) V₁ = I × 4 = 8 V  ·  V₂ = I × 6 = 12 V  (check: 8 + 12 = 20 V ✓)

V1 = 8 V, V2 = 12 V

Both branches see the full 10 V.

(a) I₁ = 10 ÷ 5 = 2 A  ·  I₂ = 10 ÷ 20 = 0.5 A

(b) I_total = 2 + 0.5 = 2.5 A

(c) 1/R_p = 1/5 + 1/20 = 4/20 + 1/20 = 5/20, so R_p = 20/5 = 4 Ω  (check: I = V/R_p = 10/4 = 2.5 A ✓)

I1 = 2 A, I2 = 0.5 A, Rp = 4 Ω

R_p = (2 × 2) / (2 + 2) = 4/4 = 1 Ω

R_total = 1 + 1 = 2 Ω

I = 9 ÷ 2 = 4.5 A

V_s = 4.5 × 1 = 4.5 V  ·  V_p = 4.5 × 1 = 4.5 V  (check: 4.5 + 4.5 = 9 V ✓)

Since both parallel resistors are equal, the current splits evenly: I₁ = I₂ = 4.5 ÷ 2 = 2.25 A

I = 4.5 A, I1 = I2 = 2.25 A

Predict: with the 1.0 Ω branch cut, all the current that used to split between two branches now has only one road left — the 2.5 Ω branch — so its current should increase. The voltmeter is wired straight across the battery terminals and draws no current, so it keeps reading the full EMF no matter what the branches are doing: it should stay at 12 V.

Check: with the 1.0 Ω branch open, current only flows through the 1.0 Ω series resistor and the 2.5 Ω branch: R_total = 1.0 + 2.5 = 3.5 Ω.

I = 12 ÷ 3.5 ≈ 3.4 A — up from 2.0 A before the fault, confirming the prediction.

2.5 Ω branch: 2.0 A → 3.4 A. Voltmeter: stays at 12 V.

Go to the lab and use the "Top branch" switch under Break the Circuit to see this exact scenario animate.

Rearrange I = Q/t to make t the subject: t = Q / I.

t = 20 ÷ 4 = 5 s

V = W / Q = 150 ÷ 25 = 6 V

R = V / I = 18 ÷ 3 = 6 Ω

R_total = 1 + 2 + 3 = 6 Ω

I = 12 ÷ 6 = 2 A — the same everywhere in the loop

V₁ = 2 × 1 = 2 V  ·  V₂ = 2 × 2 = 4 V  ·  V₃ = 2 × 3 = 6 V  (check: 2 + 4 + 6 = 12 V ✓)

I = 2 A, V1 = 2 V, V2 = 4 V, V3 = 6 V

All three branches see the full 8 V.

I₁ = 8 ÷ 2 = 4 A  ·  I₂ = 8 ÷ 4 = 2 A  ·  I₃ = 8 ÷ 4 = 2 A

I_total = 4 + 2 + 2 = 8 A

1/R_p = 1/2 + 1/4 + 1/4 = 2/4 + 1/4 + 1/4 = 4/4 = 1, so R_p = 1 Ω  (check: I = 8/1 = 8 A ✓)

I1 = 4 A, I2 = I3 = 2 A, Itotal = 8 A, Rp = 1 Ω

R_p = (3 × 6) / (3 + 6) = 18/9 = 2 Ω

R_total = 2 + 2 = 4 Ω

I = 20 ÷ 4 = 5 A

V_s = 5 × 2 = 10 V  ·  V_p = 5 × 2 = 10 V  (check: 10 + 10 = 20 V ✓)

I₁ = 10 / 3 ≈ 3.3 A  ·  I₂ = 10 / 6 ≈ 1.7 A  (check: 3.3 + 1.7 = 5 A ✓)

I = 5 A, I1 ≈ 3.3 A, I2 ≈ 1.7 A

R_p = (4 × 12) / (4 + 12) = 48/16 = 3 Ω

R_total = 3 + 3 = 6 Ω

I = 24 ÷ 6 = 4 A

V_s = 4 × 3 = 12 V  ·  V_p = 4 × 3 = 12 V  (check: 12 + 12 = 24 V ✓)

I₁ = 12 / 4 = 3 A  ·  I₂ = 12 / 12 = 1 A  (check: 3 + 1 = 4 A ✓)

I = 4 A, I1 = 3 A, I2 = 1 A

Total resistance increases — series resistances simply add up, so an extra resistor always adds to the total.

Since I = V/R and R has increased while the battery's EMF hasn't changed, the current decreases.

Combined resistance decreases — every extra branch gives the charge another route through, and 1/R_p = 1/R₁ + 1/R₂ + … only ever grows as you add terms.

Since I = V/R_p and R_p has fallen while the EMF hasn't changed, the total current increases.

Predict: the 1.0 Ω branch now carries everything the battery supplies, so its current should rise. But because 1.0 Ω is a much easier road than the 2.5 Ω branch from Q10, expect a bigger total current here than in Q10's case.

Check: R_total = 1.0 (series) + 1.0 (only remaining branch) = 2.0 Ω.

I = 12 ÷ 2.0 = 6.0 A — up from 5.0 A before the fault (all of it now through the 1.0 Ω branch).

Compare with Q10: cutting the 2.5 Ω branch there left total current at 3.4 A. Cutting the smaller-resistance branch here leaves a bigger total current (6.0 A) — losing the easier road costs you more current than losing the harder one.

1.0 Ω branch: 5.0 A → 6.0 A. Total current: 7.0 A → 6.0 A.