IB Physics Quiz: Understand Current And Circuits

20 questions · exam conditions

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Understand Current And CircuitsQuestion 1 of 20

A cylindrical wire made of a material with resistivity ρ has length L and radius r, giving it a resistance R. This wire is replaced by a new wire of the same material with length 2L and radius r/2. What is the resistance of the new wire?

16R

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IB Physics Quiz

IB Physics Quiz: Understand Current And Circuits

Practice Understand Current And Circuits in IB Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Understand Current And Circuits, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

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All questions

Question 1

2R
4R
8R (correct answer)
16R

Explanation: Resistance is given by the formula R = ρL/A, where A is the cross-sectional area. The area is A = πr². The original resistance is R = ρL/(πr²). The new wire has length L' = 2L and radius r' = r/2. The new area is A' = π(r/2)² = πr²/4 = A/4. The new resistance is R' = ρL'/A' = ρ(2L)/(A/4) = 8(ρL/A) = 8R.

Question 2

A battery has an electromotive force (emf) of 12 V and an internal resistance of 2.0 Ω. It is connected to an external resistor of 4.0 Ω. What is the power dissipated as heat inside the battery?

8.0 W (correct answer)
16 W
24 W
36 W

Explanation: First, calculate the total resistance of the circuit: R_total = R_external + r_internal = 4.0 Ω + 2.0 Ω = 6.0 Ω. Next, find the current flowing in the circuit using Ohm's law with the emf: I = ε / R_total = 12 V / 6.0 Ω = 2.0 A. The power dissipated within the battery is due to its internal resistance: P_internal = I²r = (2.0 A)² × 2.0 Ω = 4.0 × 2.0 = 8.0 W.

Question 3

Two identical lamps are connected in series to a power supply of emf ε. A third, identical lamp is connected to an identical power supply. Assume the internal resistance of the supplies is negligible. What is the ratio of the power dissipated by one lamp in the series circuit to the power dissipated by the third lamp?

1/4 (correct answer)
1/2
1
2

Explanation: Let the resistance of each lamp be R. The third lamp is connected to a supply ε, so the power it dissipates is P_third = ε²/R. For the two lamps in series, the total resistance is R + R = 2R. The current flowing through the series circuit is I_series = ε / (2R). The power dissipated by one lamp in this circuit is P_one_series = (I_series)² × R = (ε / (2R))² × R = (ε² / (4R²)) × R = ε² / (4R). The required ratio is P_one_series / P_third = (ε² / (4R)) / (ε² / R) = 1/4.

Question 4

A heating element is made from a wire of resistance R and dissipates power P when connected to a fixed voltage supply. The wire is then cut into two equal halves, and these two halves are connected in parallel to the same voltage supply. What is the new total power dissipated?

P/2
P
2P
4P (correct answer)

Explanation: The original power is P = V²/R. When the wire is cut in half, each half has a resistance of R/2. When these two halves are connected in parallel, the equivalent resistance R_eq is found by 1/R_eq = 1/(R/2) + 1/(R/2) = 2/R + 2/R = 4/R. Therefore, R_eq = R/4. The new power dissipated by this parallel combination is P_new = V²/R_eq = V²/(R/4) = 4(V²/R) = 4P.

Question 5

A lamp rated '60 W, 120 V' and another rated '100 W, 120 V' are connected in series to a 120 V power supply. Assuming their resistances remain constant, what is the total power dissipated in the two lamps?

38 W (correct answer)
80 W
120 W
160 W

Explanation: First, calculate the resistance of each lamp from its power rating using P = V²/R. For the 60 W lamp, R₁ = (120 V)² / 60 W = 14400 / 60 = 240 Ω. For the 100 W lamp, R₂ = (120 V)² / 100 W = 14400 / 100 = 144 Ω. When connected in series, the total resistance is R_total = R₁ + R₂ = 240 Ω + 144 Ω = 384 Ω. Now, calculate the total power dissipated by this series combination using the 120 V supply: P_total = V² / R_total = (120 V)² / 384 Ω = 14400 / 384 = 37.5 W. This is approximately 38 W.

Question 6

An electric motor operates with an efficiency of 80% to lift a 200 kg mass through a vertical height of 5.0 m in 20 s. Electrical energy costs $0.30 per kWh. What is the approximate cost of this operation? (Use g = 10 N kg⁻¹).

$0.00010
$0.00083
$0.00104 (correct answer)
$0.02100

Explanation:

Calculate useful work output (gain in GPE): W_out = mgh = 200 kg × 10 N/kg × 5.0 m = 10000 J. 2. Calculate useful power output: P_out = W_out / t = 10000 J / 20 s = 500 W. 3. Calculate electrical power input using efficiency: P_in = P_out / η = 500 W / 0.80 = 625 W. 4. Calculate total electrical energy input in joules: E_in = P_in × t = 625 W × 20 s = 12500 J. 5. Convert energy to kWh: E_in(kWh) = 12500 J / (3.6 × 10⁶ J/kWh) ≈ 0.00347 kWh. 6. Calculate cost: Cost = Energy(kWh) × Price($/kWh) = 0.00347 kWh × $0.30/kWh ≈ $0.00104.

Question 7

A cylindrical conductor of resistance R is melted down and reformed into a new cylindrical conductor with half the original length. The volume of the material is conserved. Which of the following are correct consequences of this change? I. The cross-sectional area is doubled. II. The resistivity of the material is halved. III. The new resistance is R/4.

I and III only (correct answer)
II and III only
I only
III only

Explanation: I. The volume V = AL is constant. If the new length L' = L/2, then the new area A' must be 2A to keep the volume A'L' = (2A)(L/2) = AL constant. So, statement I is correct. II. Resistivity (ρ) is an intrinsic property of the material and does not change when the shape is changed. So, statement II is incorrect. III. The original resistance is R = ρL/A. The new resistance is R' = ρL'/A' = ρ(L/2)/(2A) = (1/4)(ρL/A) = R/4. So, statement III is correct. Therefore, statements I and III are correct.

Question 8

A potential difference of 12 V is applied across a resistor. During a time interval of 3.0 s, a total of 6.0 × 10¹⁹ electrons pass through it. What is the energy dissipated in the resistor during this time? (The elementary charge e = 1.6 × 10⁻¹⁹ C)

38 J
115 J (correct answer)
230 J
346 J

Explanation: First, calculate the total charge (Δq) that passes through the resistor: Δq = number of electrons × e = (6.0 × 10¹⁹) × (1.6 × 10⁻¹⁹ C) = 9.6 C. The energy dissipated (E) is the work done on the charge, given by E = V × Δq. E = 12 V × 9.6 C = 115.2 J. Distractor D (346 J) comes from incorrectly multiplying by time again (V × q × t). Distractor A (38 J) comes from calculating the power (P = Vq/t = 115.2/3 = 38.4W) and reporting it as energy.

Question 9

A thermistor, whose resistance decreases as temperature increases, is connected in series with a fixed resistor R to a constant voltage supply.

What are the effects on the current in the circuit and the potential difference across the fixed resistor R when the thermistor is heated?

Current decreases; potential difference across R decreases.
Current decreases; potential difference across R increases.
Current increases; potential difference across R decreases.
Current increases; potential difference across R increases. (correct answer)

Explanation: When the thermistor is heated, its resistance (R_T) decreases. Since it is in series with a fixed resistor (R), the total circuit resistance (R_total = R_T + R) decreases. According to Ohm's law, with a constant supply voltage V, the total current in the circuit (I = V / R_total) must increase. The potential difference across the fixed resistor is given by V_R = I × R. Since the current I increases and R is constant, the potential difference V_R must also increase.

Question 10

A copper wire (wire 1) of cross-sectional area A carries a current I, resulting in an electron drift velocity v. A second copper wire (wire 2) has a cross-sectional area of A/2 and carries a current of 2I. What is the electron drift velocity in wire 2?

v/2
v
2v
4v (correct answer)

Explanation: The formula for drift velocity is I = nAvq, where n is the number density of charge carriers, A is the cross-sectional area, v is the drift velocity, and q is the charge of a carrier. Solving for v gives v = I / (nAq). For wire 1, v₁ = I / (nAq). For wire 2, the current is I₂ = 2I and the area is A₂ = A/2. Since both wires are copper, n and q are the same. The drift velocity in wire 2 is v₂ = I₂ / (nA₂q) = (2I) / (n(A/2)q) = 4 * (I / (nAq)) = 4v₁.

Question 11

A battery with significant internal resistance is connected to an external resistor R. An identical resistor is then connected in parallel with R. What is the effect on the current flowing through the original resistor R?

It increases because the total resistance of the circuit decreases.
It decreases because the terminal potential difference of the battery decreases. (correct answer)
It remains the same because the resistance of R is unchanged.
It increases because the total current from the battery increases.

Explanation: When a second identical resistor is connected in parallel, the total external resistance decreases (from R to R/2). This causes the total resistance of the circuit (R_ext + r_int) to decrease. Consequently, the total current drawn from the battery (I_total = ε / R_total) increases. The terminal potential difference across the battery is V_terminal = ε - I_total × r_int. Since I_total increases, the 'lost volts' (I_total × r_int) increase, and therefore the terminal potential difference V_terminal decreases. The current through the original resistor R is given by I_R = V_terminal / R. Since V_terminal decreases and R is constant, the current I_R must decrease.

Question 12

A battery with an emf of 12 V and negligible internal resistance is connected to a 2.0 Ω resistor in series with a parallel combination of a 3.0 Ω resistor and a 6.0 Ω resistor. What is the potential difference across the 3.0 Ω resistor?

4.0 V
4.5 V
6.0 V (correct answer)
8.0 V

Explanation: First, find the equivalent resistance of the parallel part: 1/R_p = 1/3.0 + 1/6.0 = 2/6.0 + 1/6.0 = 3/6.0, so R_p = 2.0 Ω. Next, find the total resistance of the circuit: R_total = R_series + R_p = 2.0 Ω + 2.0 Ω = 4.0 Ω. Calculate the total current from the battery: I_total = V / R_total = 12 V / 4.0 Ω = 3.0 A. This current flows through the parallel combination. The potential difference across the parallel combination is V_p = I_total × R_p = 3.0 A × 2.0 Ω = 6.0 V. Since resistors in parallel have the same potential difference across them, the pd across the 3.0 Ω resistor is 6.0 V.

Question 13

A potential divider circuit is constructed using a light-dependent resistor (LDR) and a fixed resistor, R, connected in series to a power supply. The potential difference is measured across the fixed resistor R.

The LDR is initially in a brightly lit environment. If the LDR is then covered to place it in darkness, what is the effect on the potential difference across the fixed resistor R?

It increases, because the current in the circuit increases.
It decreases, because the total resistance of the circuit decreases.
It remains the same, because the resistance of R is fixed.
It decreases, because the current in the circuit decreases. (correct answer)

Explanation: When the LDR is moved into darkness, its resistance increases significantly. Since the LDR and fixed resistor R are in series, the total resistance of the circuit (R_total = R_LDR + R) increases. According to Ohm's law (I = V / R_total), the total current in the circuit decreases. The potential difference across the fixed resistor is given by V_R = I × R. Since I decreases and R is constant, the potential difference V_R must decrease.

Question 14

A student measures the current through a resistor as $I = 0.25 \pm 0.02 \, \text{A}$ and the voltage across it as $V = 6.0 \pm 0.3 \, \text{V}$ . Using $R = V/I$ , what is the resistance value with its absolute uncertainty?

$R = 24 \pm 2 \, \Omega$ using standard error propagation for division (correct answer)
$R = 24 \pm 3 \, \Omega$ using maximum possible error analysis
$R = 24 \pm 1 \, \Omega$ using simplified uncertainty estimation methods
$R = 24 \pm 4 \, \Omega$ using conservative uncertainty combination rules

Explanation: For division R = V/I, the fractional uncertainty is: δR/R = √[(δV/V)² + (δI/I)²] = √[(0.3/6.0)² + (0.02/0.25)²] = √[0.0025 + 0.0064] = √0.0089 = 0.094. Therefore δR = 0.094 × 24 = 2.3 ≈ 2 Ω. Choice B uses simple addition of fractional uncertainties. Choice C underestimates by using only the larger fractional uncertainty. Choice D overestimates by using worst-case addition.

Question 15

Two identical light bulbs are connected in parallel to a battery. A third identical bulb is then connected in series with the parallel combination. Compared to the initial brightness of the parallel bulbs, how does their brightness change?

Each parallel bulb becomes dimmer because the series bulb reduces available voltage across the parallel combination (correct answer)
Each parallel bulb becomes brighter because total circuit current increases with the additional bulb
Each parallel bulb maintains the same brightness because parallel voltage remains constant
Each parallel bulb brightness depends on the battery's internal resistance characteristics

Explanation: Adding the series bulb increases total circuit resistance, reducing total current. The series bulb creates a voltage drop, leaving less voltage across the parallel combination. Since P = V²/R for each bulb, reduced voltage means reduced power and dimmer bulbs. Choice B incorrectly assumes total current increases. Choice C ignores the voltage division created by the series bulb. Choice D introduces irrelevant complexity about internal resistance.

Question 16

In a Wheatstone bridge circuit, the galvanometer shows zero current when the bridge is balanced. If one of the known resistors is replaced with a resistor having twice the resistance, what adjustment must be made to rebalance the bridge?

The variable resistor must be doubled to maintain the required ratio for balance
The variable resistor must be halved to compensate for the increased reference resistance
The variable resistor must be increased by a factor of four to account for bridge sensitivity
The variable resistor adjustment depends on which specific arm contains the doubled resistor (correct answer)

Explanation: In a Wheatstone bridge, balance occurs when R₁/R₂ = R₃/R₄. The required adjustment depends on which resistor is doubled. If R₁ is doubled, then R₃ must be doubled. If R₂ is doubled, then R₃ must be halved. The location matters because the ratios must be maintained across opposite arms. Choices A, B, and C assume a specific configuration without considering that 'known resistor' could refer to different positions in the bridge.

Question 17

A capacitor in an RC circuit is initially uncharged. When connected to a DC source through a resistor, it begins charging. At the moment when the capacitor voltage equals half the source voltage, what fraction of the final stored energy has been accumulated in the capacitor?

Three-quarters of the final energy, accounting for the exponential charging curve
One-half of the final energy, since voltage is directly proportional to energy
One-quarter of the final energy, since energy depends quadratically on voltage (correct answer)
One-eighth of the final energy, considering both voltage and current dependencies

Explanation: When analyzing RC circuits, remember that energy storage in capacitors depends on the square of the voltage, not linearly on voltage itself. The energy stored in a capacitor is given by

U = \frac{1}{2}CV^2

. When the capacitor is fully charged, its voltage equals the source voltage

V_0

, so the final energy is

U_{final} = \frac{1}{2}CV_0^2

. At the moment described in the question, the capacitor voltage is

\frac{V_0}{2}

, so the current energy is

U_{current} = \frac{1}{2}C(\frac{V_0}{2})^2 = \frac{1}{2}C \cdot \frac{V_0^2}{4} = \frac{1}{4} \cdot \frac{1}{2}CV_0^2 = \frac{1}{4}U_{final}

. Answer A incorrectly assumes some complex relationship involving the exponential charging curve, but energy calculation is straightforward once you know the voltage. Answer B falls into the common trap of thinking energy is proportional to voltage—this would be true if energy were

U = CV

, but it's actually

U = \frac{1}{2}CV^2

. Answer D overcomplicates the problem by bringing in current dependencies, but stored energy depends only on the capacitor's voltage at any given moment. The key insight is recognizing the quadratic relationship between voltage and energy. When voltage is halved, energy becomes one-quarter because of the

V^2

dependence. Study tip: Always remember that capacitor energy scales as

V^2

, not

V

. This quadratic relationship appears frequently in IB Physics problems involving energy storage and transfers.

Question 18

A total charge of 450 C flows through a component in 3.0 minutes. What is the average number of electrons that pass through the component per second? (The elementary charge e = 1.6 × 10⁻¹⁹ C)

9.4 × 10¹⁸
1.6 × 10¹⁹ (correct answer)
2.8 × 10²¹
9.4 × 10²⁰

Explanation: First, convert the time to seconds: 3.0 minutes = 3.0 × 60 s = 180 s. Next, calculate the average current, which is the charge passing per second: I = Δq / Δt = 450 C / 180 s = 2.5 A, which is 2.5 C/s. To find the number of electrons per second, divide the charge per second by the charge of a single electron: Number per second = (2.5 C/s) / (1.6 × 10⁻¹⁹ C/electron) ≈ 1.56 × 10¹⁹ electrons/s. This rounds to 1.6 × 10¹⁹ electrons/s.

Question 19

An ammeter and a voltmeter are used to measure the current through and potential difference across a resistor. Which row correctly describes the connection of the meters and their ideal resistances?

Ammeter in series, voltmeter in parallel; Ammeter resistance is infinite, voltmeter resistance is zero.
Ammeter in parallel, voltmeter in series; Ammeter resistance is zero, voltmeter resistance is infinite.
Ammeter in parallel, voltmeter in series; Ammeter resistance is infinite, voltmeter resistance is zero.
Ammeter in series, voltmeter in parallel; Ammeter resistance is zero, voltmeter resistance is infinite. (correct answer)

Explanation: To measure the current flowing through a component, the ammeter must be placed in series with it so that the same current flows through the meter. To measure the potential difference across a component, the voltmeter must be placed in parallel with it. An ideal ammeter has zero resistance so it does not affect the current it is measuring. An ideal voltmeter has infinite resistance so that no current flows through it, thus not altering the circuit it is measuring.

Question 20

Which statement best describes the condition for a material to be 'ohmic'?

The material must be a metallic conductor.
The resistance of the material is directly proportional to its length.
The potential difference across the material is proportional to the current through it for a constant temperature. (correct answer)
The current is the same in all parts of the material when connected in a series circuit.

Explanation: Ohm's law states that for an ohmic conductor, the current is directly proportional to the potential difference across it, provided that physical conditions such as temperature are kept constant. This is equivalent to saying the ratio V/I (resistance) is constant. Choice C accurately describes this relationship. Choice A is too narrow; some non-metals can be ohmic. Choice B describes how resistance depends on length, which is different from Ohm's law. Choice D describes a property of any series circuit, not a specific material property.