IB Physics Quiz: Apply Current And Circuits
20 questions · exam conditions
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Apply Current And CircuitsQuestion 1 of 20

Two copper wires, X and Y, have the same length. The radius of wire X is twice the radius of wire Y. If the same current flows through both wires, what is the ratio of the electron drift velocity in X to the drift velocity in Y (vX/vYv_X / v_Y)?

1/4
1/2
2
4
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IB Physics Quiz

IB Physics Quiz: Apply Current And Circuits

Practice Apply 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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Question 1

Two copper wires, X and Y, have the same length. The radius of wire X is twice the radius of wire Y. If the same current flows through both wires, what is the ratio of the electron drift velocity in X to the drift velocity in Y (vX/vYv_X / v_Y)?

  1. 1/4 (correct answer)
  2. 1/2
  3. 2
  4. 4
Explanation: The formula for current related to drift velocity is I=nAqvI = nAqv, where A is the cross-sectional area. Since II, nn, and qq are the same for both wires, AvAv must be constant. The area is A=πr2A = \pi r^2. Therefore, r2vr^2 v is constant. We have rX2vX=rY2vYr_X^2 v_X = r_Y^2 v_Y. Given rX=2rYr_X = 2r_Y, we can substitute: (2rY)2vX=rY2vY(2r_Y)^2 v_X = r_Y^2 v_Y, which simplifies to 4rY2vX=rY2vY4r_Y^2 v_X = r_Y^2 v_Y. Dividing by rY2r_Y^2 gives 4vX=vY4v_X = v_Y, so the ratio vX/vY=1/4v_X / v_Y = 1/4.

Question 2

Two identical resistors are connected in series to a DC power supply with negligible internal resistance, dissipating a total power P. If the same two resistors are instead connected in parallel to the same power supply, what is the total power dissipated?

  1. P/4
  2. P/2
  3. 2P
  4. 4P (correct answer)
Explanation: Let each resistor have resistance R and the supply voltage be V. In series, the total resistance is Rseries=R+R=2RR_{series} = R + R = 2R. The power is P=Pseries=V2/Rseries=V2/(2R)P = P_{series} = V^2 / R_{series} = V^2 / (2R). In parallel, the total resistance is Rparallel=(R−1+R−1)−1=R/2R_{parallel} = (R^{-1} + R^{-1})^{-1} = R/2. The new power is Pparallel=V2/Rparallel=V2/(R/2)=2V2/RP_{parallel} = V^2 / R_{parallel} = V^2 / (R/2) = 2V^2/R. To find the relationship, we can express V2/RV^2/R from the first equation as 2P2P. Substituting this into the second equation gives Pparallel=2(2P)=4PP_{parallel} = 2(2P) = 4P.

Question 3

A metallic conductor is stretched to three times its original length, while its volume remains constant. What is the ratio of its new resistance to its original resistance?

  1. 1/3
  2. 3
  3. 9 (correct answer)
  4. 1/9
Explanation: Resistance is given by R=ρL/AR = \rho L / A. The volume is V=L×AV = L \times A, which is constant. If the new length is L′=3LL' = 3L, the new area must be A′=V/L′=V/(3L)=A/3A' = V / L' = V / (3L) = A/3 to keep the volume constant. The new resistance is R′=ρL′/A′=ρ(3L)/(A/3)=9(ρL/A)=9RR' = \rho L' / A' = \rho (3L) / (A/3) = 9 (\rho L / A) = 9R. Therefore, the ratio R′/RR' / R is 9.

Question 4

The current I in a conductor varies with time t according to the equation I=2tI = 2t, where I is in amperes and t is in seconds. What is the total charge that passes a point in the conductor between t = 0 s and t = 3 s?

  1. 3 C
  2. 6 C
  3. 9 C (correct answer)
  4. 18 C
Explanation: Current is the rate of flow of charge, I=dQ/dtI = dQ/dt. To find the total charge Q, we need to find the area under the current-time graph. Since the current increases linearly with time, the graph of I vs t is a straight line passing through the origin with a gradient of 2. The area under this graph from t=0 to t=3 is a triangle. At t=3 s, the current is I=2×3=6 AI = 2 \times 3 = 6 \, A. The area of the triangle is Area=12×base×height=12×(3 s)×(6 A)=9 CArea = \frac{1}{2} \times base \times height = \frac{1}{2} \times (3 \, s) \times (6 \, A) = 9 \, C.

Question 5

A battery of constant emf ε and internal resistance r is connected to a variable external resistor R. For which condition is the terminal potential difference across the battery closest to its emf ε?

  1. When R is much larger than r. (correct answer)
  2. When R is much smaller than r.
  3. When R is equal to r.
  4. When the current drawn is at its maximum possible value.
Explanation: The terminal potential difference V is given by V=ε−IrV = \varepsilon - Ir, where I is the current. The current is I=ε/(R+r)I = \varepsilon / (R + r). For V to be closest to ε, the term IrIr, known as the 'lost volts', must be as small as possible. This occurs when the current I is minimized. The current is minimized when the total resistance (R+r)(R + r) is maximized. This happens when the external resistance R is very large (approaches infinity).

Question 6

The resistance of a thermistor decreases significantly as its temperature increases. This thermistor is connected in series with a fixed 100 Ω resistor to a 5.0 V battery. What happens to the current in the circuit and the potential difference across the 100 Ω resistor as the thermistor is heated?

  1. Current decreases and potential difference decreases.
  2. Current decreases and potential difference increases.
  3. Current increases and potential difference decreases.
  4. Current increases and potential difference increases. (correct answer)
Explanation: When the thermistor is heated, its resistance decreases. Since it is in series with the fixed resistor, the total resistance of the circuit decreases. According to Ohm's Law (I=V/RtotalI = V/R_{total}), a decrease in total resistance will cause the current in the circuit to increase. The potential difference across the fixed 100 Ω resistor is given by V100=I×100 ΩV_{100} = I \times 100 \, \Omega. Since the current I increases, the potential difference across the fixed resistor also increases.

Question 7

A wire of material X and a wire of material Y are connected in series. Both wires have the same length and diameter. The resistivity of X is half the resistivity of Y. What is the ratio of the potential difference across X to the potential difference across Y?

  1. 1/4
  2. 1/2 (correct answer)
  3. 1
  4. 2
Explanation: The resistance R is given by R=ρL/AR = \rho L/A. Since L and A are the same for both wires, the resistance is directly proportional to the resistivity (R∝ρR \propto \rho). Therefore, RX/RY=ρX/ρY=1/2R_X / R_Y = \rho_X / \rho_Y = 1/2. In a series circuit, the current I is the same through both components. The potential difference across each is given by V=IRV = IR. The ratio of the potential differences is VX/VY=(IRX)/(IRY)=RX/RY=1/2V_X / V_Y = (IR_X) / (IR_Y) = R_X / R_Y = 1/2.

Question 8

A potential divider circuit consists of a Light Dependent Resistor (LDR) and a fixed resistor R connected in series to a constant voltage supply. The output voltage is measured across the fixed resistor R. When the circuit is taken from a brightly lit room to a dark room, how do the resistance of the LDR and the output voltage change?

  1. LDR resistance decreases and output voltage decreases.
  2. LDR resistance decreases and output voltage increases.
  3. LDR resistance increases and output voltage decreases. (correct answer)
  4. LDR resistance increases and output voltage increases.
Explanation: The resistance of an LDR increases as the light intensity decreases (i.e., when going into a dark room). The output voltage across the fixed resistor R is given by the potential divider formula: Vout=Vsupply×RR+RLDRV_{out} = V_{supply} \times \frac{R}{R + R_{LDR}}. As RLDRR_{LDR} increases, the denominator (R+RLDR)(R + R_{LDR}) increases. This causes the overall fraction to decrease, so the output voltage VoutV_{out} decreases.

Question 9

A battery with internal resistance r is connected to a variable resistor R. The power dissipated in R is measured for different values of R. For which value of R is the power dissipated in the external resistor R a maximum?

  1. When R is zero.
  2. When R is equal to r. (correct answer)
  3. When R is twice r.
  4. When R is infinitely large.
Explanation: This is a standard result known as the maximum power transfer theorem. The power dissipated in the external resistor is given by P=I2RP = I^2 R. The current is I=ε/(R+r)I = \varepsilon / (R+r). Substituting for I, we get P=(εR+r)2RP = (\frac{\varepsilon}{R+r})^2 R. To find the maximum, one can use calculus (differentiating P with respect to R and setting to zero) or analyze the function. The result is that maximum power is transferred to the external load R when the load resistance equals the internal resistance of the source, i.e., R=rR = r.

Question 10

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

  1. 4.0 V
  2. 6.0 V
  3. 8.0 V (correct answer)
  4. 9.0 V
Explanation: First, calculate the equivalent resistance of the parallel combination: Rp=(12.0+16.0)−1=(3+16.0)−1=(46.0)−1=1.5 ΩR_p = (\frac{1}{2.0} + \frac{1}{6.0})^{-1} = (\frac{3+1}{6.0})^{-1} = (\frac{4}{6.0})^{-1} = 1.5 \, \Omega. Next, calculate the total equivalent resistance of the circuit: Rtotal=Rseries+Rp=3.0 Ω+1.5 Ω=4.5 ΩR_{total} = R_{series} + R_p = 3.0 \, \Omega + 1.5 \, \Omega = 4.5 \, \Omega. Then, find the total current from the battery: I=V/Rtotal=12 V/4.5 Ω=8/3 AI = V / R_{total} = 12 \, V / 4.5 \, \Omega = 8/3 \, A. Finally, calculate the potential difference across the 3.0 Ω resistor: V3Ω=I×Rseries=(8/3 A)×3.0 Ω=8.0 VV_{3\Omega} = I \times R_{series} = (8/3 \, A) \times 3.0 \, \Omega = 8.0 \, V.

Question 11

A capacitor is being charged through a resistor. When the voltage across the capacitor reaches 63% of its maximum value, the rate of change of charge on the capacitor is 2.0×10−4 C/s2.0 \times 10^{-4} \text{ C/s}. What was the initial rate of change of charge when charging began?

  1. 5.4×10−4 C/s5.4 \times 10^{-4} \text{ C/s} (correct answer)
  2. 3.2×10−4 C/s3.2 \times 10^{-4} \text{ C/s}
  3. 7.4×10−4 C/s7.4 \times 10^{-4} \text{ C/s}
  4. 1.3×10−4 C/s1.3 \times 10^{-4} \text{ C/s}
Explanation: For RC charging, I(t)=I0e−t/RCI(t) = I_0 e^{-t/RC} where I0I_0 is the initial current. At 63% of maximum voltage (one time constant), VC=0.63VmaxV_C = 0.63V_{max}, which occurs at t=RCt = RC. At this time, I=I0e−1=0.37I0I = I_0 e^{-1} = 0.37I_0. Since I=dQdt=2.0×10−4I = \frac{dQ}{dt} = 2.0 \times 10^{-4} at this point, we have 0.37I0=2.0×10−40.37I_0 = 2.0 \times 10^{-4}, so I0=2.0×10−40.37=5.4×10−4 C/sI_0 = \frac{2.0 \times 10^{-4}}{0.37} = 5.4 \times 10^{-4} \text{ C/s}.

Question 12

A capacitor C=2.0 μFC = 2.0 \text{ μF} is charged to 12 V and then connected through a resistor R=1.0 MΩR = 1.0 \text{ MΩ} to an uncharged capacitor C2=3.0 μFC_2 = 3.0 \text{ μF}. What is the energy dissipated in the resistor during the entire redistribution process?

  1. 86 μJ (correct answer)
  2. 58 μJ
  3. 144 μJ
  4. 72 μJ
Explanation: Initial energy: Ei=12CV2=12(2.0×10−6)(12)2=144 μJE_i = \frac{1}{2}CV^2 = \frac{1}{2}(2.0 \times 10^{-6})(12)^2 = 144 \text{ μJ}. At equilibrium, both capacitors have the same voltage. By charge conservation: Qtotal=CV=(2.0×10−6)(12)=24 μCQ_{total} = CV = (2.0 \times 10^{-6})(12) = 24 \text{ μC}. Final voltage: Vf=QtotalC+C2=24×10−65.0×10−6=4.8 VV_f = \frac{Q_{total}}{C + C_2} = \frac{24 \times 10^{-6}}{5.0 \times 10^{-6}} = 4.8 \text{ V}. Final energy: Ef=12(C+C2)Vf2=12(5.0×10−6)(4.8)2=58 μJE_f = \frac{1}{2}(C + C_2)V_f^2 = \frac{1}{2}(5.0 \times 10^{-6})(4.8)^2 = 58 \text{ μJ}. Energy dissipated: Ei−Ef=144−58=86 μJE_i - E_f = 144 - 58 = 86 \text{ μJ}.

Question 13

A battery charger applies a constant current of 2.0 A to charge a 12 V car battery (internal resistance 0.05 Ω) that is initially at 10.8 V. Assuming the battery's emf increases linearly with charge at a rate of 0.002 V per coulomb, how much energy is supplied by the charger in the first 60 seconds?

  1. 1.51 kJ
  2. 1.44 kJ
  3. 1.68 kJ
  4. 1.32 kJ (correct answer)
Explanation: Charge delivered: Q=It=2.0×60=120 CQ = It = 2.0 \times 60 = 120 \text{ C}. Initial battery emf: ε0=10.8 V\varepsilon_0 = 10.8 \text{ V}. Final battery emf: εf=10.8+(0.002)(120)=11.04 V\varepsilon_f = 10.8 + (0.002)(120) = 11.04 \text{ V}. Average battery emf: εˉ=10.8+11.042=10.92 V\bar{\varepsilon} = \frac{10.8 + 11.04}{2} = 10.92 \text{ V}. Voltage drop across internal resistance: Vr=Ir=2.0×0.05=0.1 VV_r = Ir = 2.0 \times 0.05 = 0.1 \text{ V}. Average charger voltage: Vcharger=εˉ+Vr=10.92+0.1=11.02 VV_{charger} = \bar{\varepsilon} + V_r = 10.92 + 0.1 = 11.02 \text{ V}. Energy supplied: E=Vcharger×I×t=11.02×2.0×60=1.32 kJE = V_{charger} \times I \times t = 11.02 \times 2.0 \times 60 = 1.32 \text{ kJ}.

Question 14

A student measures the terminal voltage of a battery under different load conditions and plots terminal voltage versus current. The graph yields a straight line with y-intercept 9.0 V and slope -0.8 V/A. If this battery is connected to a resistor that draws 3.0 A, what percentage of the battery's total power output is delivered to the external resistor?

  1. 73.3% (correct answer)
  2. 80.0%
  3. 66.7%
  4. 90.0%
Explanation: From the graph: emf ε=9.0 V\varepsilon = 9.0 \text{ V} (y-intercept) and internal resistance r=0.8 Ωr = 0.8 \text{ Ω} (magnitude of slope). At I=3.0 AI = 3.0 \text{ A}: terminal voltage V=9.0−0.8(3.0)=6.6 VV = 9.0 - 0.8(3.0) = 6.6 \text{ V}. Total power output from battery: Ptotal=εI=9.0×3.0=27.0 WP_{total} = \varepsilon I = 9.0 \times 3.0 = 27.0 \text{ W}. Power delivered to external resistor: Pext=VI=6.6×3.0=19.8 WP_{ext} = VI = 6.6 \times 3.0 = 19.8 \text{ W}. Percentage: 19.827.0×100%=73.3%\frac{19.8}{27.0} \times 100\% = 73.3\%.

Question 15

A battery with internal resistance r=0.5 Ωr = 0.5 \text{ Ω} and emf ε=12 V\varepsilon = 12 \text{ V} is connected to a circuit containing two resistors: R1=3.0 ΩR_1 = 3.0 \text{ Ω} in series with a parallel combination of R2=4.0 ΩR_2 = 4.0 \text{ Ω} and R3=6.0 ΩR_3 = 6.0 \text{ Ω}. What is the current through R3R_3?

  1. 1.8 A
  2. 2.4 A
  3. 0.96 A (correct answer)
  4. 1.2 A
Explanation: First, find the equivalent resistance of the parallel combination: 1R23=14.0+16.0=512\frac{1}{R_{23}} = \frac{1}{4.0} + \frac{1}{6.0} = \frac{5}{12}, so R23=2.4 ΩR_{23} = 2.4 \text{ Ω}. Total circuit resistance is Rtotal=r+R1+R23=0.5+3.0+2.4=5.9 ΩR_{total} = r + R_1 + R_{23} = 0.5 + 3.0 + 2.4 = 5.9 \text{ Ω}. Total current is I=εRtotal=125.9=2.03 AI = \frac{\varepsilon}{R_{total}} = \frac{12}{5.9} = 2.03 \text{ A}. Voltage across the parallel combination is V23=I×R23=2.03×2.4=4.88 VV_{23} = I \times R_{23} = 2.03 \times 2.4 = 4.88 \text{ V}. Current through R3R_3 is I3=V23R3=4.886.0=0.96 AI_3 = \frac{V_{23}}{R_3} = \frac{4.88}{6.0} = 0.96 \text{ A}.

Question 16

Two identical resistors RR are connected in parallel, and this combination is connected in series with a third resistor 2R2R and a battery of emf ε\varepsilon. If one of the parallel resistors fails (becomes an open circuit), by what factor does the total power dissipated by the circuit change?

  1. Increases by a factor of 1.5
  2. Decreases by a factor of 1.5 (correct answer)
  3. Increases by a factor of 2.25
  4. Decreases by a factor of 2.25
Explanation: Initially: parallel combination has resistance R/2R/2, total resistance is Rtotal1=2R+R/2=5R/2R_{total1} = 2R + R/2 = 5R/2. Power is P1=ε2Rtotal1=2ε25RP_1 = \frac{\varepsilon^2}{R_{total1}} = \frac{2\varepsilon^2}{5R}. After failure: total resistance becomes Rtotal2=2R+R=3RR_{total2} = 2R + R = 3R. New power is P2=ε23RP_2 = \frac{\varepsilon^2}{3R}. The ratio is P2P1=ε2/3R2ε2/5R=56=0.67\frac{P_2}{P_1} = \frac{\varepsilon^2/3R}{2\varepsilon^2/5R} = \frac{5}{6} = 0.67, so power decreases by a factor of 10.67=1.5\frac{1}{0.67} = 1.5.

Question 17

An electric kettle is rated 2.4 kW at 240 V. It is protected by a fuse. Which of the following is the most suitable fuse rating?

  1. 1 A
  2. 5 A
  3. 10 A
  4. 13 A (correct answer)
Explanation: First, calculate the normal operating current of the kettle using the power rating. Power P=IVP = IV, so current I=P/VI = P/V. The power is P=2.4 kW=2400 WP = 2.4 \, \text{kW} = 2400 \, W, and the voltage is V=240 VV = 240 \, V. So, I=2400 W/240 V=10 AI = 2400 \, W / 240 \, V = 10 \, A. A fuse must have a rating slightly higher than the normal operating current to allow for minor fluctuations but to blow under fault conditions. Of the choices given, 13 A is the next standard rating above 10 A and is therefore the most suitable.

Question 18

A potential divider uses a 10 kΩ resistor and a 20 kΩ resistor in series with a 6.0 V supply of negligible internal resistance. What is the output voltage if it is taken across the 10 kΩ resistor?

  1. 1.5 V
  2. 2.0 V (correct answer)
  3. 3.0 V
  4. 4.0 V
Explanation: Using the potential divider formula, Vout=Vin×R1R1+R2V_{out} = V_{in} \times \frac{R_1}{R_1 + R_2}, where R1R_1 is the resistor across which the output is taken. Here, Vin=6.0 VV_{in} = 6.0 \, V, R1=10 kΩR_1 = 10 \, k\Omega, and R2=20 kΩR_2 = 20 \, k\Omega. The total resistance is Rtotal=10 kΩ+20 kΩ=30 kΩR_{total} = 10 \, k\Omega + 20 \, k\Omega = 30 \, k\Omega. So, Vout=6.0 V×10 kΩ30 kΩ=6.0 V×13=2.0 VV_{out} = 6.0 \, V \times \frac{10 \, k\Omega}{30 \, k\Omega} = 6.0 \, V \times \frac{1}{3} = 2.0 \, V.

Question 19

An electron is accelerated from rest through a potential difference of 500 V. What is the work done on the electron?

  1. 500 J
  2. 8.0 x 10⁻¹⁷ J (correct answer)
  3. 3.1 x 10¹⁸ J
  4. 1.6 x 10⁻¹⁹ J
Explanation: The work done W on a charge q when it moves through a potential difference V is given by the equation W=qVW = qV. The charge of an electron is the elementary charge, e=1.60×10−19 Ce = 1.60 \times 10^{-19} \, C. The potential difference is V=500 VV = 500 \, V. Therefore, the work done is W=(1.60×10−19 C)×(500 V)=8.0×10−17 JW = (1.60 \times 10^{-19} \, C) \times (500 \, V) = 8.0 \times 10^{-17} \, J.

Question 20

A battery has an emf of 12 V and an internal resistance r. When connected to a 4.0 Ω resistor, the current is 2.0 A. What is the power dissipated as heat within the battery?

  1. 4.0 W
  2. 8.0 W (correct answer)
  3. 16 W
  4. 24 W
Explanation: First, find the internal resistance r using the formula ε=I(R+r)\varepsilon = I(R + r). Rearranging gives r=ε/I−Rr = \varepsilon/I - R. Substituting the values: r=(12 V/2.0 A)−4.0 Ω=6.0 Ω−4.0 Ω=2.0 Ωr = (12 \, V / 2.0 \, A) - 4.0 \, \Omega = 6.0 \, \Omega - 4.0 \, \Omega = 2.0 \, \Omega. The power dissipated within the battery (in the internal resistance) is given by Pinternal=I2rP_{internal} = I^2 r. So, P=(2.0 A)2×2.0 Ω=4.0×2.0=8.0 WP = (2.0 \, A)^2 \times 2.0 \, \Omega = 4.0 \times 2.0 = 8.0 \, W.