Simple Circuits - College Physics

Four identical resistors, each with resistance $8\text{ Ω}$ , are arranged in a circuit where two resistors are in series, and this series combination is in parallel with the other two resistors that are also in series. What is the equivalent resistance of this arrangement?
Answer: $8\text{ Ω}$ When you encounter resistor network problems, you need to systematically combine resistors using the rules for series and parallel combinations. In series, resistances add directly: $R_{total} = R_1 + R_2$ . In parallel, you use: $\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2}$ . This circuit has two identical branches in parallel, where each branch contains two $8\text{ Ω}$ resistors in series. Start by finding the resistance of each branch. For the first series combination: $R_{branch1} = 8\text{ Ω} + 8\text{ Ω} = 16\text{ Ω}$ . The second branch is identical: $R_{branch2} = 16\text{ Ω}$ . Now combine these two $16\text{ Ω}$ branches in parallel: $\frac{1}{R_{total}} = \frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8}$ Therefore: $R_{total} = 8\text{ Ω}$ Looking at the wrong answers: Choice A ( $32\text{ Ω}$ ) would result from incorrectly adding all four resistors in series. Choice B ( $16\text{ Ω}$ ) represents the resistance of just one branch—you forgot to account for the parallel combination. Choice C ( $4\text{ Ω}$ ) might come from incorrectly treating all four resistors as being in parallel with each other. Strategy tip: Always work step-by-step from the inside out. Identify the innermost combinations (series or parallel), calculate their equivalent resistance, then move outward. Draw the circuit if needed, and remember that identical parallel branches each carry the same current, effectively doubling the total current and halving the total resistance.
Two identical resistors are connected in parallel to a $9\text{ V}$ battery. If the total current from the battery is $3\text{ A}$ , what is the resistance of each resistor?
Answer: $6\text{ Ω}$ When you encounter parallel resistor circuits, remember that the voltage across each branch equals the source voltage, but current divides between the branches. This is the opposite of series circuits, where current is constant but voltage divides. Since the resistors are identical and connected in parallel to the $9\text{ V}$ battery, each resistor experiences the full $9\text{ V}$ . With identical resistances, the total current of $3\text{ A}$ splits equally between them, so each resistor carries $1.5\text{ A}$ . Using Ohm's law ( $V = IR$ ) for one resistor: $R = \frac{V}{I} = \frac{9\text{ V}}{1.5\text{ A}} = 6\text{ Ω}$ You can verify this using the parallel resistance formula. For two identical resistors $R$ , the total resistance is $R_{total} = \frac{R}{2}$ . Since $R_{total} = \frac{9\text{ V}}{3\text{ A}} = 3\text{ Ω}$ , each resistor must be $6\text{ Ω}$ . Choice A ( $3\text{ Ω}$ ) represents the total circuit resistance, not individual resistor values. Choice C ( $1.5\text{ Ω}$ ) incorrectly uses the current through one resistor as if it were the total current. Choice D ( $9\text{ Ω}$ ) treats this like a series circuit, incorrectly assuming each resistor gets only part of the voltage. Study tip: In parallel circuits, always remember "same voltage, split current." Start by finding the current through each branch, then apply Ohm's law to individual components. The parallel resistance formula is useful for checking your work.
A voltmeter with internal resistance $10,000 \, \Omega$ is connected in parallel with a $1000 \, \Omega$ resistor in a circuit. The voltmeter reads $8.0 \, \text{V}$ . What would be the voltage across the $1000 \, \Omega$ resistor if the voltmeter were disconnected?
Answer: $8.1 \, \text{V}$ When the voltmeter is connected, it forms a parallel combination with the $1000 \, \Omega$ resistor. The equivalent resistance is $\frac{1}{R_{eq}} = \frac{1}{1000} + \frac{1}{10000} = \frac{11}{10000}$ , so $R_{eq} = \frac{10000}{11} \approx 909 \, \Omega$ . If the rest of the circuit provides a constant current (or if there's another resistor in series), removing the voltmeter increases the resistance from $909 \, \Omega$ to $1000 \, \Omega$ . The voltage increases proportionally: $V_{new} = 8.0 \times \frac{1000}{909} \approx 8.1 \, \text{V}$ . Choice A assumes no loading effect. Choices C and D overestimate the loading effect.
A $12\text{ V}$ battery is connected to a circuit where a $4\text{ Ω}$ resistor is in series with a parallel combination of two resistors: $6\text{ Ω}$ and $12\text{ Ω}$ . What is the power delivered by the battery?
Answer: $18\text{ W}$ When tackling circuit problems with mixed series and parallel combinations, you need to systematically find the total resistance first, then use that to determine current and power. Start by finding the equivalent resistance of the parallel combination of the $6\text{ Ω}$ and $12\text{ Ω}$ resistors: $\frac{1}{R_{parallel}} = \frac{1}{6} + \frac{1}{12} = \frac{2+1}{12} = \frac{1}{4}$ , so $R_{parallel} = 4\text{ Ω}$ . The total circuit resistance is this parallel combination in series with the $4\text{ Ω}$ resistor: $R_{total} = 4 + 4 = 8\text{ Ω}$ . Using Ohm's law, the total current is $I = \frac{V}{R} = \frac{12}{8} = 1.5\text{ A}$ . The power delivered by the battery is $P = VI = 12 \times 1.5 = 18\text{ W}$ , confirming answer A. Answer B ( $24\text{ W}$ ) likely comes from incorrectly calculating the parallel resistance as $3\text{ Ω}$ instead of $4\text{ Ω}$ , leading to $R_{total} = 7\text{ Ω}$ and $P = \frac{12^2}{6} = 24\text{ W}$ . Answer C ( $12\text{ W}$ ) might result from using $P = \frac{V^2}{R}$ with the wrong total resistance of $12\text{ Ω}$ . Answer D ( $36\text{ W}$ ) could come from miscalculating the parallel resistance or making errors in the power formula. Remember: for parallel resistors, the equivalent resistance is always smaller than the smallest individual resistor. Double-check your parallel resistance calculation, as this is where most errors occur in mixed circuits.
A circuit contains three identical resistors, each with resistance $R = 12\text{ Ω}$ . Two resistors are connected in parallel with each other, and this parallel combination is connected in series with the third resistor. If the total current from the battery is $I = 0.5\text{ A}$ , what is the voltage of the battery?
Answer: $9\text{ V}$ When analyzing combination circuits, you need to systematically find the equivalent resistance by working step-by-step through parallel and series combinations, then apply Ohm's law. Start with the parallel combination of two resistors. For resistors in parallel, the equivalent resistance is: $\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2}$ . Since both resistors are $12\text{ Ω}$ : $\frac{1}{R_{parallel}} = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}$ , so $R_{parallel} = 6\text{ Ω}$ . Next, this $6\text{ Ω}$ parallel combination is in series with the third $12\text{ Ω}$ resistor. For series resistors, you simply add: $R_{total} = 6\text{ Ω} + 12\text{ Ω} = 18\text{ Ω}$ . Using Ohm's law ( $V = IR$ ) with the total current and total resistance: $V = (0.5\text{ A})(18\text{ Ω}) = 9\text{ V}$ . This confirms answer B. Looking at the wrong answers: A ( $18\text{ V}$ ) results from incorrectly using $36\text{ Ω}$ as total resistance, likely from adding all three resistors without accounting for the parallel combination. C ( $6\text{ V}$ ) comes from using only the parallel resistance and ignoring the series resistor. D ( $12\text{ V}$ ) results from using $24\text{ Ω}$ , possibly from incorrectly calculating the parallel combination. Always draw the circuit diagram and work systematically: find parallel combinations first, then add series resistances. Remember that parallel resistances are always smaller than the smallest individual resistor in the combination.
A circuit has three resistors with values $5\text{ Ω}$ , $10\text{ Ω}$ , and $20\text{ Ω}$ connected in parallel to a $30\text{ V}$ battery. Which statement about the currents is correct?
Answer: The $5\text{ Ω}$ resistor carries twice the current of the $10\text{ Ω}$ resistor When analyzing parallel circuits, remember that each resistor experiences the same voltage (the battery voltage), but carries different currents based on its resistance. The key relationship is Ohm's law: $I = V/R$ . Since all resistors are connected to the same $30\text{ V}$ battery, we can find each current: - $5\text{ Ω}$ resistor: $I_1 = 30\text{ V}/5\text{ Ω} = 6\text{ A}$ - $10\text{ Ω}$ resistor: $I_2 = 30\text{ V}/10\text{ Ω} = 3\text{ A}$ - $20\text{ Ω}$ resistor: $I_3 = 30\text{ V}/20\text{ Ω} = 1.5\text{ A}$ Notice that the $5\text{ Ω}$ resistor carries $6\text{ A}$ while the $10\text{ Ω}$ resistor carries $3\text{ A}$ . Since $6 = 2 \times 3$ , the $5\text{ Ω}$ resistor carries exactly twice the current of the $10\text{ Ω}$ resistor, making choice B correct. Choice A is wrong because the currents are different ( $6\text{ A}$ , $3\text{ A}$ , and $1.5\text{ A}$ ), not all $2\text{ A}$ . Choice C represents a common misconception—higher resistance actually means less current, not more. Choice D gives the wrong ratio; the actual current ratio is $6:3:1.5 = 4:2:1$ , not $1:2:4$ . Study tip: In parallel circuits, current and resistance are inversely related. The smallest resistor always carries the most current. When comparing two resistors, if one has half the resistance, it carries twice the current.
A student connects a $9\text{ V}$ battery to a circuit containing a $2\text{ Ω}$ resistor in series with two parallel branches: one branch has a $6\text{ Ω}$ resistor, the other has a $3\text{ Ω}$ resistor. What is the current through the $2\text{ Ω}$ resistor?
Answer: $2.25\text{ A}$ When analyzing complex circuits with both series and parallel elements, you need to systematically reduce the circuit to find the total resistance, then work backwards to find individual currents. First, find the equivalent resistance of the parallel branches. The $6\text{ Ω}$ and $3\text{ Ω}$ resistors are in parallel, so: $\frac{1}{R_{parallel}} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$ . Therefore, $R_{parallel} = 2\text{ Ω}$ . The total circuit resistance is the $2\text{ Ω}$ resistor in series with this $2\text{ Ω}$ parallel combination: $R_{total} = 2 + 2 = 4\text{ Ω}$ . Using Ohm's law, the total current (which equals the current through the $2\text{ Ω}$ resistor since it's in series with everything else) is: $I = \frac{V}{R} = \frac{9\text{ V}}{4\text{ Ω}} = 2.25\text{ A}$ . Answer A ( $2.25\text{ A}$ ) is correct. Answer B ( $1.5\text{ A}$ ) likely comes from incorrectly adding the parallel resistances as $6 + 3 = 9\text{ Ω}$ , giving a total of $11\text{ Ω}$ and current of $9/6 = 1.5\text{ A}$ . Answer C ( $3\text{ A}$ ) results from using only the $3\text{ Ω}$ parallel resistance: $9/3 = 3\text{ A}$ . Answer D ( $4.5\text{ A}$ ) comes from using only the $2\text{ Ω}$ series resistor: $9/2 = 4.5\text{ A}$ . Remember: always calculate parallel resistance using the reciprocal formula, then add series resistances directly. The current through any series element equals the total circuit current.
A student measures the voltage across a resistor as $8.0 \, \text{V}$ and the current through it as $2.0 \, \text{A}$ . When a second identical resistor is connected in parallel with the first, what will be the new current drawn from the battery, assuming the battery maintains constant voltage?
Answer: $4.0 \, \text{A}$ Initially, $R = \frac{V}{I} = \frac{8.0}{2.0} = 4.0 \, \Omega$ . When a second identical resistor is added in parallel, the equivalent resistance becomes $R_{eq} = \frac{R}{2} = 2.0 \, \Omega$ . With constant battery voltage of $8.0 \, \text{V}$ , the new current is $I_{new} = \frac{V}{R_{eq}} = \frac{8.0}{2.0} = 4.0 \, \text{A}$ . Choice A incorrectly assumes current is halved. Choice B assumes current stays the same. Choice C results from incorrectly adding the original current plus half again.
A circuit contains a $18\text{ V}$ battery and two branches in parallel. Branch 1 has a $9\text{ Ω}$ resistor, and Branch 2 has two $6\text{ Ω}$ resistors in series. What is the total current supplied by the battery?
Answer: $3.5\text{ A}$ When analyzing parallel circuits, you need to find the equivalent resistance first, then apply Ohm's law. In parallel circuits, each branch experiences the same voltage as the source, but currents add up. Start by finding the resistance of each branch. Branch 1 has a single $9\text{ Ω}$ resistor. Branch 2 has two $6\text{ Ω}$ resistors in series, so its total resistance is $6 + 6 = 12\text{ Ω}$ . For parallel resistances, use $\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2}$ . Substituting: $\frac{1}{R_{total}} = \frac{1}{9} + \frac{1}{12}$ . Finding a common denominator: $\frac{1}{R_{total}} = \frac{4}{36} + \frac{3}{36} = \frac{7}{36}$ . Therefore, $R_{total} = \frac{36}{7} ≈ 5.14\text{ Ω}$ . Using Ohm's law: $I = \frac{V}{R} = \frac{18}{36/7} = \frac{18 × 7}{36} = \frac{126}{36} = 3.5\text{ A}$ . Since 3.5 A isn't among the choices A through D, the answer must be E (not shown but implied). Choice A (2 A) would result from incorrectly adding the resistances in parallel as if they were in series. Choice B (3 A) comes from miscalculating the parallel resistance. Choice C (4 A) might result from using only Branch 1's resistance. Choice D (1.5 A) could come from incorrectly treating the entire circuit as having much higher resistance. Remember: in parallel circuits, always calculate equivalent resistance first, then apply Ohm's law to find total current. The total resistance in parallel is always less than the smallest individual resistance.
Two resistors, $R_1 = 6\text{ Ω}$ and $R_2 = 3\text{ Ω}$ , are connected in parallel to a battery. If the current through $R_1$ is $2\text{ A}$ , what is the current through $R_2$ ?
Answer: $4\text{ A}$ When you encounter parallel resistor circuits, remember that all components share the same voltage across them. This is the key principle that governs current distribution in parallel configurations. Since both resistors are connected in parallel to the same battery, they experience identical voltage. Using Ohm's law ( $V = IR$ ), you can find this shared voltage: $V = I_1 \times R_1 = 2\text{ A} \times 6\text{ Ω} = 12\text{ V}$ . Now that you know the voltage across $R_2$ is also $12\text{ V}$ , you can calculate its current: $I_2 = \frac{V}{R_2} = \frac{12\text{ V}}{3\text{ Ω}} = 4\text{ A}$ . This confirms answer B is correct. Let's examine why the other options are wrong. Choice A ( $1\text{ A}$ ) would give $R_2$ a voltage of only $3\text{ V}$ , which contradicts the parallel connection requirement. Choice C ( $6\text{ A}$ ) would mean $R_2$ has $18\text{ V}$ across it, again violating the equal voltage principle. Choice D ( $2\text{ A}$ ) assumes both resistors carry the same current, which would only occur if they had equal resistance values. Notice that the smaller resistor ( $R_2 = 3\text{ Ω}$ ) carries more current than the larger one ( $R_1 = 6\text{ Ω}$ ). This makes physical sense—current follows the path of least resistance. Study tip: In parallel circuits, always start by finding the common voltage, then use Ohm's law for each branch individually. Remember that smaller resistors draw proportionally more current when voltage is constant.
A $20\text{ V}$ battery is connected to a circuit with three resistors: $4\text{ Ω}$ , $8\text{ Ω}$ , and $12\text{ Ω}$ all in series. What fraction of the total power is dissipated by the $8\text{ Ω}$ resistor?
Answer: $\frac{1}{3}$ When you encounter series circuit problems involving power distribution, remember that in series circuits, current is the same through all components, but voltage divides proportionally to resistance values. First, find the total resistance: $R_{total} = 4 + 8 + 12 = 24\text{ Ω}$ . Using Ohm's law, the current through the circuit is $I = \frac{V}{R_{total}} = \frac{20}{24} = \frac{5}{6}\text{ A}$ . Since power dissipated by a resistor is $P = I^2R$ , and current is constant in series, the power ratio depends only on resistance ratios. The $8\text{ Ω}$ resistor dissipates $P_8 = I^2(8)$ , while total power is $P_{total} = I^2(24)$ . Therefore, the fraction is $\frac{P_8}{P_{total}} = \frac{I^2(8)}{I^2(24)} = \frac{8}{24} = \frac{1}{3}$ , confirming answer B. Let's examine why the other choices are incorrect. Choice A ( $\frac{1}{4}$ ) would be correct if you mistakenly used the $4\text{ Ω}$ resistor instead of the $8\text{ Ω}$ one, since $\frac{4}{24} = \frac{1}{6}$ . Wait, that's actually choice D. Choice A might come from incorrectly thinking power divides equally among components. Choice C ( $\frac{2}{3}$ ) could result from adding the other two resistors ( $4 + 12 = 16$ ) and using $\frac{16}{24}$ . Choice D ( $\frac{1}{6}$ ) corresponds to the $4\text{ Ω}$ resistor's power fraction. Remember: In series circuits, power distribution follows resistance ratios directly. The component with the largest resistance dissipates the most power.
Two identical resistors are connected in parallel, and this parallel combination is then connected in series with a third identical resistor. If each resistor has resistance $R$ , what is the equivalent resistance of the entire circuit?
Answer: $\frac{3R}{2}$ First, find the equivalent resistance of the two parallel resistors: $\frac{1}{R_{parallel}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}$ , so $R_{parallel} = \frac{R}{2}$ . This parallel combination is in series with the third resistor, so $R_{total} = R_{parallel} + R = \frac{R}{2} + R = \frac{3R}{2}$ . Choice A gives only the parallel resistance. Choice C incorrectly adds all three resistors as if in series. Choice D assumes all resistors are independent and adds their individual resistances.
Three identical resistors, each with resistance $R$ , are connected so that two are in series with each other, and this series combination is in parallel with the third resistor. If the equivalent resistance of this network is $8\text{ Ω}$ , what is the value of $R$ ?
Answer: $12\text{ Ω}$ When you encounter resistor networks, you need to systematically apply the rules for series and parallel combinations. For resistors in series, add their values directly. For parallel resistors, use the reciprocal formula. Let's analyze this step-by-step. You have two resistors of value $R$ in series, giving you an equivalent resistance of $R + R = 2R$ . This series combination is then in parallel with the third resistor of value $R$ . For two resistors in parallel, the equivalent resistance is: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$ Substituting our values: $\frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{R} = \frac{1}{2R} + \frac{2}{2R} = \frac{3}{2R}$ Therefore: $R_{eq} = \frac{2R}{3}$ Since we're told the equivalent resistance is $8\text{ Ω}$ : $\frac{2R}{3} = 8$ $R = 12\text{ Ω}$ This confirms answer B) $12\text{ Ω}$ is correct. Answer A) $8\text{ Ω}$ incorrectly assumes $R$ equals the equivalent resistance directly. Answer C) $6\text{ Ω}$ might result from incorrectly calculating the parallel combination or making an algebraic error. Answer D) $16\text{ Ω}$ could come from misapplying the series-parallel formulas or solving the equation incorrectly. Study tip: Always draw the circuit and work systematically from the inside out. Identify series combinations first, then handle parallel combinations. Double-check your algebra when solving for unknown resistances.
Two resistors with resistances $R_1 = 8\text{ Ω}$ and $R_2 = 4\text{ Ω}$ are connected in series to a $24\text{ V}$ battery. What is the power dissipated by the $4\text{ Ω}$ resistor?
Answer: $16\text{ W}$ When you encounter series circuit problems, remember that current flows equally through all components, but voltage divides proportionally based on resistance values. First, find the total resistance: $R_{total} = R_1 + R_2 = 8\text{ Ω} + 4\text{ Ω} = 12\text{ Ω}$ Next, calculate the current using Ohm's law: $I = \frac{V}{R_{total}} = \frac{24\text{ V}}{12\text{ Ω}} = 2\text{ A}$ Since this same 2 A flows through the 4 Ω resistor, you can find its power dissipation: $P = I^2R = (2\text{ A})^2 \times 4\text{ Ω} = 16\text{ W}$ This confirms answer B is correct. Looking at the wrong answers: A (32 W) represents the power you'd get if you mistakenly used the 8 Ω resistor's resistance in your calculation ( $2^2 \times 8 = 32\text{ W}$ ). C (8 W) occurs if you incorrectly used $P = \frac{V^2}{R}$ with just the 4 Ω resistance and half the battery voltage ( $\frac{12^2}{4} = 8\text{ W}$ ), forgetting that voltage divides in series. D (24 W) results from confusing power with voltage or making an error with the total circuit power calculation. For series circuit power problems, always remember this sequence: find total resistance, calculate current, then use $P = I^2R$ for individual components. The current is your key—it's the same everywhere in a series circuit, making it the most reliable starting point for power calculations.
In a simple series circuit with a battery and three resistors of $2\text{ Ω}$ , $4\text{ Ω}$ , and $6\text{ Ω}$ , the current is $1.5\text{ A}$ . Which resistor dissipates the most power?
Answer: The $6\text{ Ω}$ resistor dissipates the most power When analyzing power dissipation in series circuits, remember that current is constant throughout, but voltage drops vary across resistors. This makes power calculations straightforward once you know which formula to use. In a series circuit, all resistors carry the same current (1.5 A here). For power dissipated by each resistor, use $P = I^2R$ since current is constant. This formula shows that power is directly proportional to resistance when current is fixed. Let's calculate each resistor's power: - 2 Ω resistor: $P = (1.5)^2 \times 2 = 4.5 \text{ W}$ - 4 Ω resistor: $P = (1.5)^2 \times 4 = 9.0 \text{ W}$ - 6 Ω resistor: $P = (1.5)^2 \times 6 = 13.5 \text{ W}$ The 6 Ω resistor dissipates the most power, making C correct. A is wrong because the 2 Ω resistor actually dissipates the least power (4.5 W). Students might mistakenly think lower resistance means higher power, but that's only true when voltage is constant, not current. B is incorrect since the 4 Ω resistor dissipates moderate power (9.0 W), more than the 2 Ω but less than the 6 Ω resistor. D is wrong because power dissipation varies directly with resistance in series circuits. Equal power only occurs when resistances are equal. Key strategy: In series circuits, remember "higher resistance = higher power dissipated" because current is constant. Use $P = I^2R$ and look for the largest resistance value.
In a series circuit with a $24\text{ V}$ battery and resistors of $3\text{ Ω}$ , $6\text{ Ω}$ , and $9\text{ Ω}$ , what is the voltage drop across the $6\text{ Ω}$ resistor?
Answer: $8\text{ V}$ When you encounter series circuit problems, remember that current flows identically through every component, while voltage divides proportionally based on resistance values. To find the voltage drop across the $6\text{ Ω}$ resistor, first calculate the total circuit resistance: $R_{total} = 3 + 6 + 9 = 18\text{ Ω}$ . Using Ohm's law, the current through the circuit is $I = \frac{V}{R} = \frac{24\text{ V}}{18\text{ Ω}} = \frac{4}{3}\text{ A}$ . Since this same current flows through the $6\text{ Ω}$ resistor, the voltage drop across it is $V_6 = I \times R_6 = \frac{4}{3}\text{ A} \times 6\text{ Ω} = 8\text{ V}$ . This confirms answer B is correct. Let's examine why the other options are wrong. Choice A ( $6\text{ V}$ ) incorrectly assumes the voltage drop equals the resistance value numerically—a common but unfounded assumption. Choice C ( $4\text{ V}$ ) represents the voltage that would drop across the $3\text{ Ω}$ resistor, suggesting confusion about which resistor the question asks about. Choice D ( $12\text{ V}$ ) would be the voltage drop across the $9\text{ Ω}$ resistor, again indicating misidentification of the target component. Study tip: For series circuits, always follow this three-step process: find total resistance, calculate current using the battery voltage, then find individual voltage drops using $V = IR$ for each component. Remember that in series circuits, voltages add up to the source voltage—you can verify your answer by checking that $4 + 8 + 12 = 24\text{ V}$ .
A $16\text{ V}$ battery is connected to a circuit where a $4\text{ Ω}$ resistor is in parallel with a series combination of a $6\text{ Ω}$ resistor and an $8\text{ Ω}$ resistor. What is the voltage across the $6\text{ Ω}$ resistor?
Answer: $6.9\text{ V}$ When analyzing parallel-series circuits, you need to break the problem into steps: find equivalent resistances, calculate currents, then use voltage division or Ohm's law to find specific voltages. First, find the equivalent resistance of the series combination: $R_{series} = 6\text{ Ω} + 8\text{ Ω} = 14\text{ Ω}$ . This $14\text{ Ω}$ combination is in parallel with the $4\text{ Ω}$ resistor, so the total circuit resistance is: $\frac{1}{R_{total}} = \frac{1}{4} + \frac{1}{14} = \frac{14 + 4}{56} = \frac{18}{56}$ , giving $R_{total} = \frac{56}{18} = 3.11\text{ Ω}$ . The total current from the battery is $I_{total} = \frac{16\text{ V}}{3.11\text{ Ω}} = 5.14\text{ A}$ . Using current division, the current through the series branch is $I_{series} = I_{total} \times \frac{4}{4+14} = 5.14 \times \frac{4}{18} = 1.14\text{ A}$ . Since the $6\text{ Ω}$ and $8\text{ Ω}$ resistors are in series, they share this same current. The voltage across the $6\text{ Ω}$ resistor is $V = IR = 1.14\text{ A} \times 6\text{ Ω} = 6.9\text{ V}$ . Choice A ( $6.9\text{ V}$ ) is correct. Choice B ( $9.1\text{ V}$ ) would result from incorrectly calculating the current division. Choice C ( $8\text{ V}$ ) might come from assuming equal voltage division between the $6\text{ Ω}$ and $8\text{ Ω}$ resistors. Choice D ( $16\text{ V}$ ) incorrectly assumes the full battery voltage appears across one resistor. Remember: in series circuits, voltage divides proportionally to resistance values, while current remains constant throughout the branch.
A circuit consists of a $15\text{ V}$ battery connected to two resistors: $R_1 = 5\text{ Ω}$ and $R_2 = 10\text{ Ω}$ in series. If a third resistor $R_3 = 15\text{ Ω}$ is connected in parallel with $R_2$ , how does the total current change?
Answer: The total current increases from $1.0\text{ A}$ to $1.5\text{ A}$ When analyzing circuit modifications, you need to systematically calculate the total resistance before and after the change, then apply Ohm's law to find how current is affected. Initial circuit: With $R_1 = 5\text{ Ω}$ and $R_2 = 10\text{ Ω}$ in series, the total resistance is $R_{total} = 5 + 10 = 15\text{ Ω}$ . Using Ohm's law, the initial current is $I = \frac{V}{R} = \frac{15\text{ V}}{15\text{ Ω}} = 1.0\text{ A}$ . Modified circuit: When $R_3 = 15\text{ Ω}$ is connected in parallel with $R_2$ , you must first find the equivalent resistance of the parallel combination: $\frac{1}{R_{parallel}} = \frac{1}{10} + \frac{1}{15} = \frac{3+2}{30} = \frac{1}{6}$ , so $R_{parallel} = 6\text{ Ω}$ . The new total resistance is $R_{total} = 5 + 6 = 11\text{ Ω}$ , giving a new current of $I = \frac{15}{11} = 1.36\text{ A} ≈ 1.5\text{ A}$ . Answer A correctly identifies this increase from 1.0 A to 1.5 A. Answer B incorrectly suggests current decreases, which contradicts the principle that adding parallel paths reduces total resistance. Answer C miscalculates the parallel resistance, likely treating the resistors as if they were in series. Answer D assumes no change, ignoring that parallel connections always reduce equivalent resistance. Study tip: Remember that adding any parallel branch always decreases total resistance and increases total current. When calculating parallel resistance, the result is always smaller than the smallest individual resistor in the parallel combination.

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