Resistance, Resistivity, and Ohm's Law - College Physics

A wire carrying a current of $2.5 \, A$ has a voltage drop of $7.5 \, V$ across a $1.5 \, m$ length. If the wire has a circular cross-section with radius $0.5 \, mm$ , what is the resistivity of the wire material?
Answer: $1.6 \times 10^{-6} \, \Omega \cdot m$ When you encounter a problem involving current, voltage, and wire dimensions, you're dealing with electrical resistance and resistivity. Resistivity is a material property that tells you how strongly a substance opposes electric current flow, regardless of the object's shape or size. To find resistivity, you need to combine Ohm's law with the resistance formula for a wire. First, use Ohm's law ( $V = IR$ ) to find the resistance: $R = V/I = 7.5 \, V / 2.5 \, A = 3.0 \, \Omega$ . Next, apply the wire resistance formula: $R = \rho L/A$ , where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area. For a circular wire, $A = \pi r^2 = \pi (0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7} \, m^2$ . Solving for resistivity: $\rho = RA/L = (3.0)(7.85 \times 10^{-7})/1.5 = 1.57 \times 10^{-6} \, \Omega \cdot m$ , which rounds to $1.6 \times 10^{-6} \, \Omega \cdot m$ . Choice A is correct. Choice B ( $2.4 \times 10^{-6}$ ) likely results from using diameter instead of radius when calculating area. Choice C ( $3.1 \times 10^{-6}$ ) suggests forgetting to divide by length or making an area calculation error. Choice D ( $4.7 \times 10^{-6}$ ) combines multiple errors, possibly using diameter and incorrect unit conversions. Remember: always convert units to meters before calculating, and use radius (not diameter) for circular cross-sections. The two-step approach—find resistance first, then resistivity—prevents calculation errors.
A student measures the voltage across a resistor as 12.0 V and the current through it as 3.00 A. The student then connects a second identical resistor in parallel with the first. If the voltage source maintains constant voltage, what will be the new current reading through the original resistor?
Answer: 3.00 A because the voltage across the original resistor remains unchanged When analyzing parallel resistor circuits, the key insight is that parallel branches maintain the same voltage across each component, regardless of how many branches you add. Let's work through this systematically. Initially, you have one resistor with 12.0 V across it and 3.00 A through it. Using Ohm's law, this resistor has a resistance of $R = V/I = 12.0\text{ V}/3.00\text{ A} = 4.00\text{ Ω}$ . When you add an identical resistor in parallel, each resistor still experiences the full 12.0 V from the voltage source. Since the original resistor's resistance (4.00 Ω) and the voltage across it (12.0 V) both remain unchanged, the current through it must also remain 3.00 A according to Ohm's law. Now let's examine why the other answers miss the mark. Choice A incorrectly assumes the original current gets divided between the resistors, but this confuses series behavior with parallel behavior. Choice B correctly notes that total circuit resistance halves, but incorrectly applies this to find current through one resistor rather than total current from the source. Choice C recognizes that total circuit current doubles (which it does, from 3.00 A to 6.00 A), but wrongly suggests this affects the current through the original resistor. Study tip: Remember that in parallel circuits, voltage is the same across each branch, while currents add up. Each parallel branch operates independently - adding more branches doesn't change the current through existing branches, only the total current drawn from the source.
A variable resistor is adjusted so that its resistance varies according to $R(t) = R_0(1 + \alpha t)$ where $\alpha > 0$ is a constant. If this resistor is connected to a constant voltage source $V$ , how does the power dissipated in the resistor change with time?
Answer: Power decreases as $P(t) = \frac{V^2}{R_0(1 + \alpha t)}$ due to inverse relationship with resistance When analyzing circuits with time-varying components, you need to apply basic circuit laws while carefully tracking how each quantity changes with time. Since power dissipated in a resistor is given by $P = \frac{V^2}{R}$ , and the voltage source is constant while resistance varies as $R(t) = R_0(1 + \alpha t)$ , you can substitute directly: $P(t) = \frac{V^2}{R_0(1 + \alpha t)}$ . As time increases and $\alpha > 0$ , the denominator grows larger, making the power smaller. This inverse relationship means power decreases over time, confirming answer A. Option B incorrectly assumes a direct relationship between resistance and power. While resistance does increase linearly, power depends on the inverse of resistance, so it actually decreases. Option C falls into the trap of confusing constant voltage with constant power. A constant voltage source maintains fixed voltage across its terminals, but the power it delivers depends on the circuit's resistance—as resistance changes, so does power. Option D incorrectly suggests exponential behavior and misunderstands how voltage and resistance interact. The voltage remains constant, and there's no compounding effect that would create exponential growth. Study tip: Remember that $P = \frac{V^2}{R}$ shows power is inversely proportional to resistance when voltage is constant. When you see time-varying circuit elements, substitute the time-dependent expressions into the basic power formulas and analyze the mathematical relationship—don't rely on intuition about what "should" happen to power.
A variable resistor is adjusted so that when $15 \, V$ is applied across it, the current is $0.75 \, A$ . If the resistor is then adjusted to triple its resistance value, what current will flow when the same $15 \, V$ is applied?
Answer: $0.25 \, A$ When you encounter problems involving variable resistors and changing resistance values, you're working with Ohm's Law: $V = IR$ , where voltage, current, and resistance are directly related. First, let's find the initial resistance. With $V = 15 \, V$ and $I = 0.75 \, A$ , we can solve for resistance: $R = \frac{V}{I} = \frac{15}{0.75} = 20 \, \Omega$ . When the resistor is adjusted to triple its resistance, the new resistance becomes $R_{new} = 3 \times 20 = 60 \, \Omega$ . Since the same voltage ( $15 \, V$ ) is applied, we can find the new current: $I_{new} = \frac{V}{R_{new}} = \frac{15}{60} = 0.25 \, A$ . Looking at the answer choices: Choice A ( $0.25 \, A$ ) is correct—this is exactly what we calculated. Choice B ( $0.50 \, A$ ) represents doubling the resistance instead of tripling it, a common calculation error. Choice C ( $0.75 \, A$ ) is the original current, which would only be correct if resistance remained unchanged. Choice D ( $1.5 \, A$ ) incorrectly assumes that tripling resistance triples current, showing a fundamental misunderstanding of Ohm's Law. Remember the key relationship: current and resistance are inversely proportional when voltage is constant. If resistance increases by a factor of 3, current decreases by the same factor. This inverse relationship is crucial for solving circuit problems efficiently—when one goes up, the other must go down proportionally.
A battery with internal resistance $r$ is connected to an external load resistance $R$ . For what value of $R$ is the power dissipated in the external resistor maximized?
Answer: $R = r$ to match the internal and external resistances This is a classic maximum power transfer problem that appears frequently in circuit analysis. When you see a battery with internal resistance connected to a load, think about how the current and power depend on both resistances. To find when power is maximized in the external resistor, start with the circuit current: $I = \frac{\varepsilon}{R + r}$ , where $\varepsilon$ is the battery's EMF. The power dissipated in the external resistor is $P = I^2R = \frac{\varepsilon^2 R}{(R + r)^2}$ . To maximize this power, take the derivative with respect to $R$ and set it equal to zero: $\frac{dP}{dR} = \frac{\varepsilon^2[(R + r)^2 - R \cdot 2(R + r)]}{(R + r)^4} = 0$ . This simplifies to $(R + r) - 2R = 0$ , giving us $R = r$ . The maximum power transfer occurs when the load resistance equals the internal resistance. Answer A ( $R = r/2$ ) incorrectly focuses on minimizing total resistance, but minimum resistance doesn't maximize power in the load. Answer B ( $R = 2r$ ) has no theoretical basis—having the external resistance "dominate" doesn't optimize power transfer. Answer D ( $R \rightarrow \infty$ ) maximizes voltage across the load but minimizes current, resulting in very low power since $P = VI$ and current approaches zero. Remember the maximum power transfer theorem: for any circuit with a fixed internal resistance, maximum power is delivered to the load when the load resistance equals the source's internal resistance. This $R = r$ condition appears in many electrical engineering applications.
A heating element is designed to operate at 1200 W when connected to 120 V. Due to thermal expansion, its resistance increases by 15% when hot. What is the actual power consumption when the element reaches its operating temperature?
Answer: 1044 W because the resistance increase reduces current flow through the element When analyzing heating elements and resistance changes, you need to understand how power, voltage, current, and resistance interact through Ohm's law and the power equation. Start by finding the cold resistance using $P = \frac{V^2}{R}$ . With 1200 W at 120 V: $R_{cold} = \frac{V^2}{P} = \frac{(120)^2}{1200} = 12 \text{ ohms}$ When the element heats up, its resistance increases by 15%: $R_{hot} = 12 \times 1.15 = 13.8 \text{ ohms}$ The actual power at operating temperature becomes: $P_{actual} = \frac{V^2}{R_{hot}} = \frac{(120)^2}{13.8} = 1044 \text{ W}$ Option A incorrectly suggests a proportional decrease (15% resistance increase = 15% power decrease), but the relationship between resistance and power is inverse, not linear. Option B makes the fundamental error of thinking resistance increases boost power consumption—this ignores that higher resistance actually restricts current flow. Option C assumes the heating element somehow maintains constant power output, but real heating elements are passive resistors that follow Ohm's law; they don't actively regulate their power consumption. Option D correctly recognizes that increased resistance reduces current flow ( $I = \frac{V}{R}$ ), which in turn reduces power consumption since $P = VI$ . Study tip: For heating element problems, always remember that resistance typically increases with temperature for most materials, and higher resistance means lower power consumption at constant voltage. Calculate the cold resistance first, then apply the temperature effect.
A $12 \, V$ battery is connected across a resistor, and a current of $0.75 \, A$ flows through the circuit. If the battery is replaced with a $9 \, V$ battery and the same resistor is used, what current will flow through the circuit?
Answer: $0.56 \, A$ This question tests your understanding of Ohm's Law, which states that voltage, current, and resistance are related by $V = IR$ . When you see a circuit problem involving a battery replacement, think about how changing one variable affects the others while keeping resistance constant. First, let's find the resistance using the initial conditions. With $V = 12 \, V$ and $I = 0.75 \, A$ , we can solve for resistance: $R = \frac{V}{I} = \frac{12}{0.75} = 16 \, \Omega$ . This resistance remains the same when we swap batteries. Now with the $9 \, V$ battery and the same $16 \, \Omega$ resistor, we can find the new current: $I = \frac{V}{R} = \frac{9}{16} = 0.5625 \, A$ , which rounds to $0.56 \, A$ . Looking at the wrong answers: Choice B ( $0.60 \, A$ ) might result from incorrectly assuming the resistance is $15 \, \Omega$ instead of $16 \, \Omega$ . Choice C ( $0.67 \, A$ ) could come from using $\frac{9}{13.5}$ , perhaps from a calculation error in finding resistance. Choice D ( $0.75 \, A$ ) represents the trap of thinking current stays constant regardless of voltage—this ignores Ohm's Law entirely. The correct answer is A) $0.56 \, A$ . Study tip: In Ohm's Law problems, always identify what stays constant. Here, resistance is fixed while voltage changes, so current must change proportionally. Practice calculating each variable systematically: find the unknown quantity first, then apply it to the new conditions.
A carbon resistor has a resistance of $1000 \, \Omega$ at $20°C$ . If the temperature coefficient of resistance for carbon is $-5.0 \times 10^{-4} \, /°C$ , what is the resistance at $-30°C$ ?
Answer: $1025 \, \Omega$ When you encounter temperature coefficient problems, you're dealing with how material properties change with temperature. The key relationship is: $R = R_0[1 + \alpha(T - T_0)]$ , where $R_0$ is the initial resistance, $\alpha$ is the temperature coefficient, and $(T - T_0)$ is the temperature change. Here, you start with $R_0 = 1000 \, \Omega$ at $T_0 = 20°C$ , and need to find the resistance at $T = -30°C$ . The temperature change is $\Delta T = -30°C - 20°C = -50°C$ . Notice that carbon has a negative temperature coefficient ( $\alpha = -5.0 \times 10^{-4} \, /°C$ ), meaning its resistance increases as temperature decreases. Substituting into the formula: $R = 1000[1 + (-5.0 \times 10^{-4})(-50)] = 1000[1 + 0.025] = 1000 \times 1.025 = 1025 \, \Omega$ Looking at the wrong answers: Choice A ( $975 \, \Omega$ ) represents what you'd get if you forgot that carbon's temperature coefficient is negative—treating it as positive would decrease resistance with decreasing temperature. Choice B ( $1000 \, \Omega$ ) ignores the temperature effect entirely. Choice D ( $1050 \, \Omega$ ) likely comes from calculation errors, possibly doubling the temperature coefficient effect. Remember that negative temperature coefficients (common in semiconductors and carbon) mean resistance increases as temperature decreases. Always pay attention to the sign of both the temperature coefficient and the temperature change—getting either wrong will lead you to an incorrect answer.
Two wires are made of the same material but have different dimensions. Wire A has length $L$ and cross-sectional area $A$ . Wire B has length $2L$ and cross-sectional area $A/2$ . What is the ratio of the resistance of wire B to the resistance of wire A?
Answer: $4:1$ When you encounter resistance problems involving different wire dimensions, always start with the fundamental resistance formula: $R = \rho \frac{L}{A}$ , where $\rho$ is resistivity (a material property), $L$ is length, and $A$ is cross-sectional area. For Wire A: $R_A = \rho \frac{L}{A}$ For Wire B: $R_B = \rho \frac{2L}{A/2} = \rho \frac{2L \cdot 2}{A} = \rho \frac{4L}{A}$ The ratio of resistances is: $\frac{R_B}{R_A} = \frac{\rho \frac{4L}{A}}{\rho \frac{L}{A}} = \frac{4L}{A} \cdot \frac{A}{L} = 4$ This gives us a 4:1 ratio, confirming answer C. Let's examine why the other options are incorrect: A) 1:1 would only be true if both length and area changes exactly canceled out, but doubling length while halving area creates a net 4× increase in resistance. B) 2:1 incorrectly assumes only the length change matters (2L vs L) while ignoring the area reduction, or vice versa. D) 1:2 gets the ratio backwards and significantly underestimates the effect of the combined dimensional changes. Study tip: Remember that resistance increases with length and decreases with area. When both changes work in the same direction (longer length AND smaller area both increase resistance), multiply their individual effects. Here, 2× length increase and 2× area decrease combine to give 2 × 2 = 4× total resistance increase.
Two wires made of the same material have the same resistance. Wire A has twice the length of wire B. If wire A has cross-sectional area $A_A$ and wire B has cross-sectional area $A_B$ , what is the ratio $A_A/A_B$ ?
Answer: $2$ because the area must compensate for the doubled length Using $R = \rho L/A$ where $\rho$ is the same for both wires. For equal resistances: $\frac{\rho L_A}{A_A} = \frac{\rho L_B}{A_B}$ . Since $L_A = 2L_B$ , we have $\frac{2L_B}{A_A} = \frac{L_B}{A_B}$ , which gives $\frac{2}{A_A} = \frac{1}{A_B}$ , so $A_A = 2A_B$ or $A_A/A_B = 2$ . The area must be twice as large to compensate for the doubled length. Choice A inverts the relationship. Choice B ignores the geometric differences. Choice D incorrectly squares the length factor.
A resistor dissipates $36 \, W$ when connected to a $12 \, V$ source. If the same resistor is connected to a $6 \, V$ source, what power does it dissipate?
Answer: $9.0 \, W$ When you encounter power dissipation problems with resistors, remember that power depends on both voltage and resistance, and there are multiple equivalent formulas to use depending on what information you have. First, let's find the resistance of this resistor. Using $P = \frac{V^2}{R}$ , we can solve for $R$ : $36 = \frac{12^2}{R} = \frac{144}{R}$ $R = \frac{144}{36} = 4 \, \Omega$ Now that we know the resistance is $4 \, \Omega$ , we can find the power dissipated when connected to a $6 \, V$ source: $P = \frac{V^2}{R} = \frac{6^2}{4} = \frac{36}{4} = 9 \, W$ Looking at the wrong answers: Choice A ( $4.5 \, W$ ) comes from incorrectly assuming power scales linearly with voltage—dividing the original power by 2 since voltage was halved. Choice C ( $18 \, W$ ) results from mistakenly thinking power is directly proportional to voltage rather than voltage squared. Choice D ( $24 \, W$ ) likely comes from calculation errors or misapplying formulas. The key insight is that power is proportional to voltage squared ( $P \propto V^2$ ) when resistance is constant. Since voltage was halved from $12 \, V$ to $6 \, V$ , the power becomes $(1/2)^2 = 1/4$ of the original: $36 \times \frac{1}{4} = 9 \, W$ . Study tip: Remember the three power formulas— $P = VI$ , $P = I^2R$ , and $P = \frac{V^2}{R}$ —and always identify which variables remain constant in the problem to choose the most efficient approach.
Two copper wires have the same length but different diameters. Wire X has diameter $d$ and wire Y has diameter $2d$ . If wire X has a resistance of $16 \, \Omega$ , what is the resistance of wire Y?
Answer: $4 \, \Omega$ When you encounter wire resistance problems, remember that resistance depends on the material's resistivity, length, and cross-sectional area according to $R = \rho \frac{L}{A}$ , where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area. Since both wires are copper with the same length, only the cross-sectional area differs. The area of a circular wire is $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$ . Wire X has area $A_X = \frac{\pi d^2}{4}$ , while wire Y has area $A_Y = \frac{\pi (2d)^2}{4} = \frac{\pi \cdot 4d^2}{4} = \pi d^2$ . Therefore, wire Y has four times the cross-sectional area of wire X. Since resistance is inversely proportional to area, wire Y has one-fourth the resistance of wire X: $R_Y = \frac{R_X}{4} = \frac{16\,\Omega}{4} = 4\,\Omega$ . This confirms answer (B) is correct. Looking at the wrong answers: (A) $2\,\Omega$ suggests the resistance decreases by a factor of 8, which would require the diameter to increase by $2\sqrt{2}$ , not 2. (C) $8\,\Omega$ implies the resistance only halves, corresponding to doubling the area rather than quadrupling it. (D) $32\,\Omega$ incorrectly suggests resistance increases with diameter, showing a fundamental misunderstanding of the inverse relationship. Study tip: Remember that doubling the diameter quadruples the area (since area depends on $d^2$ ), which quarters the resistance. Watch for this $d^2$ relationship in wire problems.
A heating element is designed to operate at $120 \, V$ and draw $10 \, A$ of current. If the voltage is reduced to $100 \, V$ and the resistance remains constant, what is the new power consumption?
Answer: $833 \, W$ This question tests your understanding of electrical power relationships and how changes in voltage affect power consumption when resistance remains constant. First, you need to find the resistance of the heating element using the original conditions. With Ohm's law, $R = \frac{V}{I} = \frac{120 \, V}{10 \, A} = 12 \, \Omega$ . Since the problem states resistance remains constant, this value stays the same when voltage changes. When voltage drops to $100 \, V$ , you can find the new power using $P = \frac{V^2}{R}$ . Substituting: $P = \frac{(100 \, V)^2}{12 \, \Omega} = \frac{10000}{12} = 833 \, W$ . This confirms answer (B). Let's examine why the other answers are incorrect. (A) 694 W might result from calculation errors or using an incorrect resistance value. (C) 1000 W represents a common trap—students might assume power scales linearly with voltage, calculating $1200 \times \frac{100}{120} = 1000$ , but power actually depends on voltage squared. (D) 1200 W is the original power consumption ( $P = VI = 120 \times 10 = 1200 \, W$ ), which some students might choose if they forget that reducing voltage decreases power. Study tip: Remember that power has multiple formulas: $P = VI$ , $P = I^2R$ , and $P = \frac{V^2}{R}$ . Choose the most convenient one based on what remains constant. When resistance is constant and voltage changes, $P = \frac{V^2}{R}$ is your best friend because it directly shows the quadratic relationship between voltage and power.
A $9.0 \, V$ battery is connected to a circuit containing a $12 \, \Omega$ resistor and an ideal ammeter in series. The ammeter reads $0.60 \, A$ . What is the internal resistance of the battery?
Answer: $3.0 \, \Omega$ When you encounter a circuit problem involving a battery with internal resistance, remember that real batteries don't provide their full emf to the external circuit. The battery's internal resistance acts like an additional resistor in series with the external components. To solve this, apply Ohm's law to the complete circuit. The battery's emf ( $9.0 \, V$ ) must equal the voltage drop across the total resistance. The total resistance includes both the external $12 \, \Omega$ resistor and the unknown internal resistance $r$ . Using Ohm's law: $V_{emf} = I \times R_{total}$ $9.0 \, V = 0.60 \, A \times (12 \, \Omega + r)$ Solving for the total resistance: $R_{total} = \frac{9.0 \, V}{0.60 \, A} = 15 \, \Omega$ Since $R_{total} = R_{external} + r$ : $15 \, \Omega = 12 \, \Omega + r$ Therefore: $r = 3.0 \, \Omega$ Choice A ( $0 \, \Omega$ ) assumes an ideal battery with no internal resistance, which would give a current of $\frac{9.0}{12} = 0.75 \, A$ . Choice C ( $6.0 \, \Omega$ ) would result from incorrectly calculating the total resistance as $18 \, \Omega$ . Choice D ( $12 \, \Omega$ ) might come from confusing the internal resistance with the external resistance. Study tip: Always remember that in real battery problems, the emf equals the current times the sum of ALL resistances in the circuit, including internal resistance. This is a fundamental concept that appears frequently in circuit analysis.
Two identical light bulbs are connected in series to a $12 \, V$ battery. Each bulb is rated at $6 \, V$ , $3 \, W$ . What is the actual power dissipated by each bulb in this series configuration?
Answer: $3.0 \, W$ When analyzing circuits with bulbs in series, you need to understand how voltage divides and how this affects power consumption compared to the bulbs' ratings. First, find each bulb's resistance using its rating: $R = \frac{V^2}{P} = \frac{6^2}{3} = 12 \, \Omega$ . In the series circuit, the total resistance is $24 \, \Omega$ , so the current is $I = \frac{12 \, V}{24 \, \Omega} = 0.5 \, A$ . Since the bulbs are identical and in series, they split the $12 \, V$ equally—each gets $6 \, V$ . The actual power dissipated by each bulb is $P = I^2R = (0.5)^2 \times 12 = 3 \, W$ , which matches their rating because they're receiving their rated voltage. Choice A ( $1.5 \, W$ ) would be incorrect if you mistakenly thought the bulbs get less than their rated voltage and calculated power as half the rating. Choice C ( $4.5 \, W$ ) might result from incorrectly assuming the bulbs somehow get more power than rated. Choice D ( $6 \, W$ ) would be wrong if you thought each bulb receives the full battery voltage, leading to $P = \frac{V^2}{R} = \frac{12^2}{12} = 12 \, W$ per bulb—but this ignores that voltage divides in series. The correct answer is B: $3.0 \, W$ . Study tip: In series circuits, always check if identical components receive their rated voltage after accounting for voltage division. When they do, they operate at their rated power. This is why two $6 \, V$ bulbs work perfectly with a $12 \, V$ source in series.
A copper wire at room temperature ( $20°C$ ) has resistance $R_0$ . If the temperature coefficient of resistivity for copper is $\alpha = 3.9 \times 10^{-3} /°C$ , what is the percentage change in resistance when the wire is heated to $120°C$ ?
Answer: $39\%$ increase because the resistance change is proportional to temperature change The resistance varies with temperature as $R = R_0[1 + \alpha(T - T_0)]$ where $\alpha$ is the temperature coefficient. The temperature change is $\Delta T = 120°C - 20°C = 100°C$ . The new resistance is $R = R_0[1 + (3.9 \times 10^{-3})(100)] = R_0[1 + 0.39] = 1.39R_0$ . The percentage change is $\frac{R - R_0}{R_0} \times 100\% = 39\%$ increase. Choice B uses an incorrect coefficient value. Choice C double-counts effects. Choice D underestimates the temperature coefficient impact.
A cylindrical wire has resistance $R$ when its length is $L$ and cross-sectional area is $A$ . If the wire is stretched to twice its original length while keeping the total volume constant, what is the new resistance in terms of $R$ ?
Answer: $4R$ When the wire is stretched to twice its length while keeping volume constant, the new length becomes $2L$ and the new cross-sectional area becomes $A/2$ (since volume $= L \cdot A$ must remain constant). Using $R = \rho L/A$ , the new resistance is $R_{new} = \rho(2L)/(A/2) = \rho \cdot 2L \cdot 2/A = 4\rho L/A = 4R$ . Choice A incorrectly accounts for only the length change. Choice C assumes the area increases rather than decreases. Choice D incorrectly inverts the relationship.
Three identical resistors, each with resistance $R$ , are connected in different configurations. Configuration A has all three in series, Configuration B has all three in parallel, and Configuration C has two in parallel connected in series with the third. What is the ratio of the total resistance of Configuration A to Configuration C?
Answer: $2:1$ When analyzing resistor networks, you need to apply the fundamental rules for series and parallel combinations systematically. For resistors in series, total resistance equals the sum of individual resistances. For parallel resistors, use $\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + ...$ Let's find each configuration's total resistance. Configuration A (all series): $R_A = R + R + R = 3R$ Configuration C requires two steps since it's a combination circuit. First, find the resistance of the two parallel resistors: $\frac{1}{R_{parallel}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}$ , so $R_{parallel} = \frac{R}{2}$ . Then add the third resistor in series: $R_C = \frac{R}{2} + R = \frac{3R}{2}$ The ratio of Configuration A to C is: $\frac{R_A}{R_C} = \frac{3R}{\frac{3R}{2}} = \frac{3R \times 2}{3R} = 2$ , giving us $2:1$ . Answer A ( $6:5$ ) likely comes from incorrectly calculating the parallel combination or making arithmetic errors. Answer B ( $3:2$ ) might result from forgetting to add the series resistor in Configuration C, comparing $3R$ to just $R$ . Answer D ( $3:1$ ) could come from mistakenly comparing Configuration A to Configuration B (all parallel), where $R_B = \frac{R}{3}$ . Remember: always work combination circuits step-by-step, solving parallel sections first, then incorporating them into the series calculation. Draw the circuit if needed to visualize the current path clearly.
Two identical resistors are connected in parallel across a battery. If one of the resistors suddenly develops an internal break (becomes an open circuit), what happens to the current through the remaining resistor?
Answer: The current doubles because all current must now flow through one path Initially, each resistor carries current $I = V/R$ where $V$ is the battery voltage. When one resistor opens, the total current that was previously split between two parallel paths ( $2V/R$ ) must now flow entirely through the remaining resistor, so the current through it becomes $2V/R = 2I$ , doubling. Choice A has the wrong reasoning about resistance. Choice C ignores that the current distribution changes. Choice D incorrectly applies series circuit reasoning to a parallel situation.
When a voltage of $6.0 \, V$ is applied across a certain resistor, the power dissipated is $18 \, W$ . What is the resistance of this resistor?
Answer: $2.0 \, \Omega$ When you encounter a problem involving voltage, power, and resistance, you're working with the fundamental relationships in electrical circuits. The key is knowing which power formula to use based on what information you're given. Since you have voltage ( $V = 6.0 \, V$ ) and power ( $P = 18 \, W$ ), the most direct approach is to use the power formula $P = \frac{V^2}{R}$ . Rearranging to solve for resistance: $R = \frac{V^2}{P}$ . Substituting the values: $R = \frac{(6.0 \, V)^2}{18 \, W} = \frac{36 \, V^2}{18 \, W} = 2.0 \, \Omega$ This confirms answer choice B is correct. Let's examine why the other options are wrong. Choice A ( $0.33 \, \Omega$ ) would result if you incorrectly calculated $\frac{P}{V^2} = \frac{18}{36}$ , essentially using the reciprocal of the correct formula. Choice C ( $3.0 \, \Omega$ ) comes from the error $\frac{V \times P}{V} = \frac{6.0 \times 18}{6.0} = 18$ , which has no physical basis. Choice D ( $12 \, \Omega$ ) results from incorrectly using $R = \frac{V \times P}{V} = \frac{6.0 \times 18}{6.0} \times 2 = 36/3$ , another computational mistake. Remember the three power formulas: $P = IV$ , $P = I^2R$ , and $P = \frac{V^2}{R}$ . Choose the one that directly uses your given variables to avoid unnecessary steps and potential errors from calculating intermediate values like current.

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