Practice — Circuit Lab

Q1 A charge of 15 C flows past a point in a wire in 5 s. Calculate the current.Easy

I = Q / t = 15 ÷ 5 = 3 A

Q2 A current of 2 A flows through a heater for 3 minutes. How much charge flows through it?Easy

Convert to seconds first: 3 minutes = 180 s.

Q = I × t = 2 × 180 = 360 C

Q3 A battery does 45 J of work moving 9 C of charge around a circuit. What is its EMF?Easy

V = W / Q = 45 ÷ 9 = 5 V

Q4 A 9 V battery moves 8 C of charge around a circuit. How much energy is transferred?Easy

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

W = 9 × 8 = 72 J

Q5 A resistor of 8 Ω has a current of 2.5 A flowing through it. Calculate the p.d. across it.Medium

V = I × R = 2.5 × 8 = 20 V

Q6 A 15 V supply drives a current of 3 A through a resistor. What is its resistance?Medium

R = V / I = 15 ÷ 3 = 5 Ω

Q7 A 4 Ω resistor and a 6 Ω resistor are connected in series to a 20 V battery. Find (a) the total resistance, (b) the current, (c) the p.d. across each resistor.Medium

(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

Q8 A 5 Ω resistor and a 20 Ω resistor are connected in parallel across a 10 V supply. Find (a) the current in each branch, (b) the total current, (c) the combined resistance.Medium

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

Q9 A 9 V battery is connected to a 1 Ω resistor in series with two parallel resistors, each 2 Ω. Find the total current, and the current through each parallel branch.Hard

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

Q10 This is the lab's circuit: 12 V battery, 1.0 Ω in series with (1.0 Ω ∥ 2.5 Ω). The 1.0 Ω branch develops a fault and goes open-circuit. Without calculating first — will the current in the 2.5 Ω branch increase, decrease, or stay the same? Will the voltmeter across the battery terminals change? Then calculate the new current to check, and verify it live in the lab.Hard · reasoning

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.