Apply Current & Circuits - IB Physics

1. 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?
   Answer: 4P Let each resistor have resistance R and the supply voltage be V. In series, the total resistance is $R_{series} = R + R = 2R$. The power is $P = P_{series} = V^2 / R_{series} = V^2 / (2R)$. In parallel, the total resistance is $R_{parallel} = (R^{-1} + R^{-1})^{-1} = R/2$. The new power is $P_{parallel} = V^2 / R_{parallel} = V^2 / (R/2) = 2V^2/R$. To find the relationship, we can express $V^2/R$ from the first equation as $2P$. Substituting this into the second equation gives $P_{parallel} = 2(2P) = 4P$.
2. An electron is accelerated from rest through a potential difference of 500 V. What is the work done on the electron?
   Answer: 8.0 x 10⁻¹⁷ J The work done W on a charge q when it moves through a potential difference V is given by the equation $W = qV$. The charge of an electron is the elementary charge, $e = 1.60 \times 10^{-19} \, C$. The potential difference is $V = 500 \, V$. Therefore, the work done is $W = (1.60 \times 10^{-19} \, C) \times (500 \, V) = 8.0 \times 10^{-17} \, J$.
3. 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?
   Answer: When R is equal to r. This is a standard result known as the maximum power transfer theorem. The power dissipated in the external resistor is given by $P = I^2 R$. The current is $I = \varepsilon / (R+r)$. Substituting for I, we get $P = (\frac{\varepsilon}{R+r})^2 R$. 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., $R = r$.
4. 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?
   Answer: 8.0 W First, find the internal resistance r using the formula $\varepsilon = I(R + r)$. Rearranging gives $r = \varepsilon/I - R$. Substituting the values: $r = (12 \, V / 2.0 \, A) - 4.0 \, \Omega = 6.0 \, \Omega - 4.0 \, \Omega = 2.0 \, \Omega$. The power dissipated within the battery (in the internal resistance) is given by $P_{internal} = I^2 r$. So, $P = (2.0 \, A)^2 \times 2.0 \, \Omega = 4.0 \times 2.0 = 8.0 \, W$.
5. 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 ($v_X / v_Y$)?
   Answer: 1/4 The formula for current related to drift velocity is $I = nAqv$, where A is the cross-sectional area. Since $I$, $n$, and $q$ are the same for both wires, $Av$ must be constant. The area is $A = \pi r^2$. Therefore, $r^2 v$ is constant. We have $r_X^2 v_X = r_Y^2 v_Y$. Given $r_X = 2r_Y$, we can substitute: $(2r_Y)^2 v_X = r_Y^2 v_Y$, which simplifies to $4r_Y^2 v_X = r_Y^2 v_Y$. Dividing by $r_Y^2$ gives $4v_X = v_Y$, so the ratio $v_X / v_Y = 1/4$.
6. A battery with internal resistance $r = 0.5 \text{ Ω}$ and emf $\varepsilon = 12 \text{ V}$ is connected to a circuit containing two resistors: $R_1 = 3.0 \text{ Ω}$ in series with a parallel combination of $R_2 = 4.0 \text{ Ω}$ and $R_3 = 6.0 \text{ Ω}$. What is the current through $R_3$?
   Answer: 0.96 A First, find the equivalent resistance of the parallel combination: $\frac{1}{R_{23}} = \frac{1}{4.0} + \frac{1}{6.0} = \frac{5}{12}$, so $R_{23} = 2.4 \text{ Ω}$. Total circuit resistance is $R_{total} = r + R_1 + R_{23} = 0.5 + 3.0 + 2.4 = 5.9 \text{ Ω}$. Total current is $I = \frac{\varepsilon}{R_{total}} = \frac{12}{5.9} = 2.03 \text{ A}$. Voltage across the parallel combination is $V_{23} = I \times R_{23} = 2.03 \times 2.4 = 4.88 \text{ V}$. Current through $R_3$ is $I_3 = \frac{V_{23}}{R_3} = \frac{4.88}{6.0} = 0.96 \text{ A}$.
7. 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?
   Answer: LDR resistance increases and output voltage decreases. 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: $V_{out} = V_{supply} \times \frac{R}{R + R_{LDR}}$. As $R_{LDR}$ increases, the denominator $(R + R_{LDR})$ increases. This causes the overall fraction to decrease, so the output voltage $V_{out}$ decreases.
8. A capacitor $C = 2.0 \text{ μF}$ is charged to 12 V and then connected through a resistor $R = 1.0 \text{ MΩ}$ to an uncharged capacitor $C_2 = 3.0 \text{ μF}$. What is the energy dissipated in the resistor during the entire redistribution process?
   Answer: 86 μJ Initial energy: $E_i = \frac{1}{2}CV^2 = \frac{1}{2}(2.0 \times 10^{-6})(12)^2 = 144 \text{ μJ}$. At equilibrium, both capacitors have the same voltage. By charge conservation: $Q_{total} = CV = (2.0 \times 10^{-6})(12) = 24 \text{ μC}$. Final voltage: $V_f = \frac{Q_{total}}{C + C_2} = \frac{24 \times 10^{-6}}{5.0 \times 10^{-6}} = 4.8 \text{ V}$. Final energy: $E_f = \frac{1}{2}(C + C_2)V_f^2 = \frac{1}{2}(5.0 \times 10^{-6})(4.8)^2 = 58 \text{ μJ}$. Energy dissipated: $E_i - E_f = 144 - 58 = 86 \text{ μJ}$.
9. 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 $2.0 \times 10^{-4} \text{ C/s}$. What was the initial rate of change of charge when charging began?
   Answer: $5.4 \times 10^{-4} \text{ C/s}$ For RC charging, $I(t) = I_0 e^{-t/RC}$ where $I_0$ is the initial current. At 63% of maximum voltage (one time constant), $V_C = 0.63V_{max}$, which occurs at $t = RC$. At this time, $I = I_0 e^{-1} = 0.37I_0$. Since $I = \frac{dQ}{dt} = 2.0 \times 10^{-4}$ at this point, we have $0.37I_0 = 2.0 \times 10^{-4}$, so $I_0 = \frac{2.0 \times 10^{-4}}{0.37} = 5.4 \times 10^{-4} \text{ C/s}$.
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?
    Answer: 8.0 V First, calculate the equivalent resistance of the parallel combination: $R_p = (\frac{1}{2.0} + \frac{1}{6.0})^{-1} = (\frac{3+1}{6.0})^{-1} = (\frac{4}{6.0})^{-1} = 1.5 \, \Omega$. Next, calculate the total equivalent resistance of the circuit: $R_{total} = R_{series} + R_p = 3.0 \, \Omega + 1.5 \, \Omega = 4.5 \, \Omega$. Then, find the total current from the battery: $I = V / R_{total} = 12 \, V / 4.5 \, \Omega = 8/3 \, A$. Finally, calculate the potential difference across the 3.0 Ω resistor: $V_{3\Omega} = I \times R_{series} = (8/3 \, A) \times 3.0 \, \Omega = 8.0 \, V$.
11. 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?
    Answer: 13 A First, calculate the normal operating current of the kettle using the power rating. Power $P = IV$, so current $I = P/V$. The power is $P = 2.4 \, \text{kW} = 2400 \, W$, and the voltage is $V = 240 \, V$. So, $I = 2400 \, W / 240 \, V = 10 \, A$. 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.
12. Two identical resistors $R$ are connected in parallel, and this combination is connected in series with a third resistor $2R$ and a battery of emf $\varepsilon$. If one of the parallel resistors fails (becomes an open circuit), by what factor does the total power dissipated by the circuit change?
    Answer: Decreases by a factor of 1.5 Initially: parallel combination has resistance $R/2$, total resistance is $R_{total1} = 2R + R/2 = 5R/2$. Power is $P_1 = \frac{\varepsilon^2}{R_{total1}} = \frac{2\varepsilon^2}{5R}$. After failure: total resistance becomes $R_{total2} = 2R + R = 3R$. New power is $P_2 = \frac{\varepsilon^2}{3R}$. The ratio is $\frac{P_2}{P_1} = \frac{\varepsilon^2/3R}{2\varepsilon^2/5R} = \frac{5}{6} = 0.67$, so power decreases by a factor of $\frac{1}{0.67} = 1.5$.
13. 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?
    Answer: Current increases and potential difference increases. 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 ($I = V/R_{total}$), 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 $V_{100} = I \times 100 \, \Omega$. Since the current I increases, the potential difference across the fixed resistor also increases.
14. 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?
    Answer: 1/2 The resistance R is given by $R = \rho L/A$. Since L and A are the same for both wires, the resistance is directly proportional to the resistivity ($R \propto \rho$). Therefore, $R_X / R_Y = \rho_X / \rho_Y = 1/2$. In a series circuit, the current I is the same through both components. The potential difference across each is given by $V = IR$. The ratio of the potential differences is $V_X / V_Y = (IR_X) / (IR_Y) = R_X / R_Y = 1/2$.
15. 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?
    Answer: 9 Resistance is given by $R = \rho L / A$. The volume is $V = L \times A$, which is constant. If the new length is $L' = 3L$, the new area must be $A' = V / L' = V / (3L) = A/3$ to keep the volume constant. The new resistance is $R' = \rho L' / A' = \rho (3L) / (A/3) = 9 (\rho L / A) = 9R$. Therefore, the ratio $R' / R$ is 9.
16. A cell is connected to an external resistor. The terminal potential difference across the cell is less than its emf. Which statement explains this?
    Answer: The cell has an internal resistance, and energy is dissipated within it. A real cell has an internal resistance, r. When current I flows from the cell, a potential difference of $Ir$ develops across this internal resistance. This potential difference, often called 'lost volts', opposes the direction of the emf. The terminal potential difference V is the emf ε minus the lost volts: $V = \varepsilon - Ir$. Therefore, when the cell is discharging (providing current), V is always less than ε due to the energy dissipated within the cell's internal resistance.
17. 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?
    Answer: 73.3% From the graph: emf $\varepsilon = 9.0 \text{ V}$ (y-intercept) and internal resistance $r = 0.8 \text{ Ω}$ (magnitude of slope). At $I = 3.0 \text{ A}$: terminal voltage $V = 9.0 - 0.8(3.0) = 6.6 \text{ V}$. Total power output from battery: $P_{total} = \varepsilon I = 9.0 \times 3.0 = 27.0 \text{ W}$. Power delivered to external resistor: $P_{ext} = VI = 6.6 \times 3.0 = 19.8 \text{ W}$. Percentage: $\frac{19.8}{27.0} \times 100\% = 73.3\%$.
18. A filament lamp is a non-ohmic device. Which statement best describes how its resistance changes as the current through it increases?
    Answer: The resistance increases because the filament gets hotter, increasing lattice vibrations. As the current through a filament lamp increases, the power dissipated as heat ($P=I^2R$) increases, causing the filament's temperature to rise significantly. For a metallic conductor like the filament, an increase in temperature leads to greater vibration of the metal lattice ions. This increases the frequency of collisions between the charge carriers (electrons) and the lattice, which impedes the flow of charge and thus increases the electrical resistance. While thermal expansion does occur, its effect on resistance is negligible compared to the effect of increased lattice vibrations.
19. A wire has a resistance of 12 Ω. It is cut into three equal pieces that are then connected in parallel with each other. What is the equivalent resistance of the parallel combination?
    Answer: 1.33 Ω When the 12 Ω wire is cut into three equal pieces, each piece will have a resistance of $12 \, \Omega / 3 = 4.0 \, \Omega$. When these three 4.0 Ω resistors are connected in parallel, the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{4.0} + \frac{1}{4.0} + \frac{1}{4.0} = \frac{3}{4.0} \, \Omega^{-1}$. Therefore, $R_{eq} = \frac{4.0}{3} \, \Omega \approx 1.33 \, \Omega$.
20. Two identical lamps are connected in parallel to a power supply with significant internal resistance. One lamp's filament breaks. What happens to the brightness of the remaining lamp?
    Answer: It becomes brighter. When the lamps are in parallel, the total external resistance is R/2. When one lamp breaks, the external resistance becomes R. The total circuit resistance increases from (R/2 + r) to (R + r). This causes the total current drawn from the supply to decrease. The 'lost volts' inside the supply are given by $Ir$. Since the total current I decreases, the lost volts decrease. The terminal potential difference across the supply, $V = \varepsilon - Ir$, therefore increases. Since the remaining lamp is connected across the terminals, the potential difference across it increases, causing it to become brighter.

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