Question 1

Use Ohm’s law for approximate value: given a source of 18 V and a current of 220 mA, estimate the resistance of the rheostat (round to the nearest practical value).

Accepted Answer

Introduction / Context: Ohm’s law (V = I * R) is the cornerstone of basic circuit calculations. When voltage and current are known, resistance follows directly. This is common when setting rheostats or selecting series resistors to achieve a target current. Given Data / Assumptions: Voltage V = 18 V. Current I = 220 mA = 0.22 A. We assume a simple DC circuit where the rheostat dominates the resistance. Concept / Approach: Solve the rearranged Ohm’s law: R = V / I. Convert milliampere to ampere to avoid unit errors. Round to a standard-ohmic value if needed for options. Step-by-Step Solution: Convert current: 220 mA = 0.22 A.Apply Ohm’s law: R = V / I = 18 / 0.22.Compute: 18 / 0.22 ≈ 81.818… Ω.Nearest listed practical value: ≈ 82 Ω. Verification / Alternative check: Back-calculate current: I = V / R = 18 / 82 ≈ 0.2195 A = 219.5 mA, which is consistent with the given 220 mA. Why Other Options Are Wrong: 8.2 kΩ and 820 Ω would yield currents far smaller than 220 mA. 8.2 Ω would give I ≈ 2.2 A, much larger than specified. Common Pitfalls: Forgetting to convert mA to A; misplacing decimals; using R = I / V instead of R = V / I. Final Answer: 82 Ω

Question 2

Apply Ohm’s law: when 12 mA flows through a 1.2 kΩ resistor, determine the voltage drop across the resistor (express in volts).

Accepted Answer

Introduction / Context: Voltage across a resistor is the product of current and resistance. This direct application of Ohm’s law appears constantly in biasing networks and sensor readouts, where small currents through kilo-ohm resistors create measurable voltages. Given Data / Assumptions: I = 12 mA = 0.012 A. R = 1.2 kΩ = 1,200 Ω. Use V = I * R. Concept / Approach: Convert units carefully to maintain consistency, then multiply. Since both values are exact in this context, no rounding beyond standard decimal arithmetic is required. Step-by-Step Solution: Convert current: 12 mA = 0.012 A.Convert resistance: 1.2 kΩ = 1,200 Ω.Compute voltage: V = I * R = 0.012 * 1,200 = 14.4 V. Verification / Alternative check: Using prefixes directly: 12 mA * 1.2 kΩ = (12 × 10^-3) * (1.2 × 10^3) = 14.4 × 10^0 = 14.4 V. Why Other Options Are Wrong: 1.4 V and 10 V do not satisfy V = I * R with the given values; 100 V is far too large for milliampere currents through kilo-ohm resistances. Common Pitfalls: Mixing milli and kilo prefixes; dropping a power of ten; forgetting that volts scale linearly with both I and R. Final Answer: 14.4 V

Question 3

Fuse-limited dimmer design: in a 120 V lamp-dimming circuit with a 3 A fuse and a lamp resistance of 20 Ω, find the minimum rheostat resistance that prevents blowing the fuse.

Accepted Answer

Introduction / Context: Series circuits with a protective fuse must ensure the total current stays at or below the fuse rating. For a lamp-dimmer using a rheostat in series with a lamp, choosing a sufficient rheostat resistance prevents overcurrent at maximum mains voltage. Given Data / Assumptions: Supply voltage: 120 V (assume DC-equivalent or RMS for calculation). Fuse rating: 3 A (maximum allowable current). Lamp resistance: 20 Ω (assumed constant for this calculation). Rheostat in series with the lamp. Concept / Approach: The largest current occurs at the smallest total series resistance. To stay within 3 A, ensure R_total ≥ V / I_max. Since R_total = R_lamp + R_rheostat_min, solve for the minimum rheostat value that satisfies the fuse limit. Step-by-Step Solution: Compute minimum total resistance: R_total_min = V / I_max = 120 / 3 = 40 Ω.Relate to components: R_total_min = R_lamp + R_rheostat_min.Solve for rheostat: R_rheostat_min = 40 − 20 = 20 Ω. Verification / Alternative check: Check current at this setting: I = V / (R_lamp + R_rheostat) = 120 / (20 + 20) = 120 / 40 = 3 A, which equals the fuse rating (no blow at steady state). Why Other Options Are Wrong: 40 Ω would be the total required resistance, not the rheostat alone. 4 Ω or 2 Ω would yield total resistance below 40 Ω, causing I > 3 A and blowing the fuse. Common Pitfalls: Forgetting that total series resistance includes both lamp and rheostat; confusing fuse current limit with power rating; ignoring that lamp resistance can vary with temperature (here treated as constant by assumption). Final Answer: 20 Ω

Question 4

Direct application of Ohm’s law: determine the source voltage required to drive 2 A through a 36 Ω resistor (show all steps clearly).

Accepted Answer

Introduction / Context: Finding the voltage for a given current through a known resistance is the most common use of Ohm’s law. This calculation is foundational in power sizing, source selection, and safety checks for resistive loads. Given Data / Assumptions: I = 2 A (current). R = 36 Ω (resistance). We assume an ideal source and purely resistive load. Concept / Approach: Use V = I * R. Multiply the given current by the resistance to get the required voltage. Verify the power as a sanity check. Step-by-Step Solution: V = I * R = 2 * 36 = 72 V.Optional power check: P = V * I = 72 * 2 = 144 W (consistent magnitude for a 36 Ω load at 2 A). Verification / Alternative check: Rearrange Ohm’s law: I = V / R → V = I * R, which we used correctly. Units: amperes * ohms = volts. Why Other Options Are Wrong: 1.8 V and 7.2 V are an order of magnitude too small; 18 V corresponds to 0.5 A through 36 Ω, not 2 A. Common Pitfalls: Accidentally dividing instead of multiplying; misreading units; ignoring that doubling current doubles voltage for a fixed resistance. Final Answer: 72 V

Question 5

Select the correct Ohm’s law rearrangement: which formula gives current I when voltage V and resistance R are known (identify the valid relationship)?

Accepted Answer

Introduction / Context: Ohm’s law relates voltage, current, and resistance. Being able to rearrange this relationship is vital when solving for a specific variable in circuit problems or lab measurements. Given Data / Assumptions: Canonical form: V = I * R. We need I in terms of V and R. Assume ideal linear resistor behavior. Concept / Approach: From V = I * R, divide both sides by R to isolate current. Ensure the units are consistent: volts divided by ohms yields amperes, confirming dimensional correctness. Step-by-Step Solution: Start with V = I * R.Divide both sides by R: V / R = I.Reorder: I = V / R. Verification / Alternative check: Unit check: V / Ω = A, which is the SI unit for current. Why Other Options Are Wrong: I = V R multiplies instead of divides; I = R / V inverts the correct ratio; V = I R is true but solves for V, not I. Common Pitfalls: Mixing up division/multiplication when isolating variables; not checking units as a safeguard. Final Answer: I = V/R

Question 6

Use resistor color code and Ohm’s law: a resistor marked red–orange–orange–silver is connected across a 50 V source; determine the current through the resistor (choose the closest standard value).

Accepted Answer

Introduction / Context: Reading resistor values from color bands and then applying Ohm’s law is a staple lab task. The 4-band code uses two significant digits, a multiplier, and a tolerance band. After decoding resistance, Ohm’s law gives the current for a given applied voltage. Given Data / Assumptions: Color code: red, orange, orange, silver. Supply: 50 V DC (or RMS equivalent for a purely resistive case). Assume nominal resistance for current calculation. Concept / Approach: Decode colors: first digit, second digit, multiplier, tolerance. Compute R, then use I = V / R. Round to the nearest provided option as typical in multiple-choice problems when exact values fall between discrete choices. Step-by-Step Solution: Decode digits: red = 2 (first), orange = 3 (second) → 23.Multiplier: orange = × 10^3 → R = 23 × 10^3 Ω = 23 kΩ.Tolerance: silver = ±10% (not needed for nominal current).Compute current: I = V / R = 50 / 23,000 ≈ 0.0021739 A ≈ 2.17 mA.Nearest choice: 2.2 mA. Verification / Alternative check: Even at ±10% tolerance, the nominal reading remains closest to 2.2 mA among options (range approx. 1.95–2.41 mA). Why Other Options Are Wrong: 2 mA is less precise than 2.2 mA to the nominal calculation; 21.4 mA and 214 mA are off by an order of magnitude or more for 23 kΩ at 50 V. Common Pitfalls: Misreading color positions; confusing the orange multiplier with × 10^2; forgetting to convert kΩ to Ω before applying V / R. Final Answer: 2.2 mA

Question 7

Compute voltage from current and resistance: if 750 µA flows through an 11 kΩ resistor, find the voltage drop across the resistor (report in volts).

Accepted Answer

Introduction / Context: Small currents through kilo-ohm resistors often appear in sensing and instrumentation circuits. Converting microamperes and kilo-ohms correctly avoids large errors when applying Ohm’s law to compute voltage drop. Given Data / Assumptions: I = 750 µA = 750 × 10^-6 A = 0.00075 A. R = 11 kΩ = 11 × 10^3 Ω = 11,000 Ω. Use V = I * R. Concept / Approach: Ensure consistent SI units before multiplying. Using exponent notation provides a quick sanity check and reduces decimal-placement errors. Step-by-Step Solution: Convert units: I = 0.00075 A; R = 11,000 Ω.Compute: V = I * R = 0.00075 * 11,000 = 8.25 V.Exponent check: (750 × 10^-6) * (11 × 10^3) = (750 * 11) × 10^-3 = 8,250 × 10^-3 = 8.25 V. Verification / Alternative check: Power check: P = V * I = 8.25 * 0.00075 ≈ 0.00619 W ≈ 6.2 mW, a reasonable dissipation for typical small-signal resistors. Why Other Options Are Wrong: 82.5 V and 146 V are too large for microampere currents; 14.6 V corresponds to different I–R values and not the given pair. Common Pitfalls: Failing to convert µA to A or kΩ to Ω; misplacing decimal points; rounding prematurely before completing the multiplication. Final Answer: 8.25 V

Question 8

Using Ohm’s law in basic circuit analysis: What is the approximate resistance setting of a rheostat if a current of 650 mA flows from a 150 V source?

Accepted Answer

Introduction / Context: This problem checks fast application of Ohm’s law to determine an unknown resistance when the source voltage and the circuit current are known. Such estimates are common when setting a rheostat to limit current to a safe value in laboratory benches and troubleshooting. Given Data / Assumptions: Source voltage V = 150 V (direct current conditions assumed). Measured current I = 650 mA = 0.65 A. Ideal lumped resistance model; temperature effects and rheostat tolerance ignored. Concept / Approach: Use Ohm’s law to relate the three variables: V = I * R. Solving for resistance gives R = V / I. Keep track of units by converting milliamperes to amperes before substitution to avoid thousandfold errors. Step-by-Step Solution: Convert current to amperes: I = 650 mA = 0.65 A.Apply Ohm’s law: R = V / I.Compute: R = 150 / 0.65 ≈ 230.77 Ω.Round to the nearest option: approximately 230 Ω. Verification / Alternative check: Back substitute to sanity-check: I = V / R ≈ 150 / 230 ≈ 0.652 A ≈ 652 mA, which is essentially the stated 650 mA once rounding is considered. The direction and magnitude are consistent. Why Other Options Are Wrong: 9.7 Ω: Would draw I ≈ 150 / 9.7 ≈ 15.5 A, far too large. 97 Ω: Gives I ≈ 1.55 A, still much larger than 0.65 A. 23 Ω: Gives I ≈ 6.5 A, an order of magnitude high. Common Pitfalls: Forgetting to convert 650 mA to 0.65 A before computing R. Rounding too early; compute first, then pick the closest option. Final Answer: 230 Ω

Question 9

Apply Ohm’s law: A 24 Ω resistor carries 4 A when connected to a source. What voltage does the source provide?

Accepted Answer

Introduction / Context: This is a direct substitution exercise with Ohm’s law, used daily to relate voltage, current, and resistance in basic circuits. Accurate substitution is essential when sizing power supplies and verifying test conditions. Given Data / Assumptions: Resistance R = 24 Ω. Current I = 4 A. Goal: source voltage V. Concept / Approach: Ohm’s law states V = I * R. Multiply current by resistance to find the terminal voltage that produces the specified current through the given resistor. Step-by-Step Solution: Identify formula: V = I * R.Insert values: V = 4 * 24.Compute: V = 96 V. Verification / Alternative check: Check power as a consistency test: P = V * I = 96 * 4 = 384 W, and also P = I^2 * R = 16 * 24 = 384 W. Both agree, confirming the result. Why Other Options Are Wrong: 960 V: Off by a factor of 10; would imply P = 3840 W. 9.6 V: Too small by a factor of 10; would give I = 0.4 A. 8 V: Would yield I = 8 / 24 = 0.333 A, not 4 A. Common Pitfalls: Mistaking decimal placement and producing a factor-of-ten error. Mixing up formulas such as R = V / I instead of solving for V directly. Final Answer: 96 V

Question 10

Estimate lamp resistance using Ohm’s law: A flashlight bulb draws 40 mA from a 4.5 V battery pack. What is the approximate resistance of the bulb filament?

Accepted Answer

Introduction / Context: Batteries and small lamps are classic examples for applying Ohm’s law. Determining filament resistance helps with estimating power and verifying that the battery can supply the load without excessive voltage sag. Given Data / Assumptions: Total battery voltage V = 4.5 V. Lamp current I = 40 mA = 0.04 A. Assume steady state at operating temperature; use linear Ohm’s law model. Concept / Approach: Compute resistance from R = V / I after converting current to amperes. The numerical accuracy is set by the significant figures of the given data, so a rounded answer is acceptable. Step-by-Step Solution: Convert current: I = 40 mA = 0.04 A.Apply Ohm’s law: R = V / I = 4.5 / 0.04.Compute: R = 112.5 Ω ≈ 112 Ω (nearest option). Verification / Alternative check: Check power: P = V * I = 4.5 * 0.04 = 0.18 W. Also P = I^2 * R ≈ 0.0016 * 112.5 ≈ 0.18 W, confirming internal consistency. Why Other Options Are Wrong: 11.2 Ω or 1.2 Ω: Would imply currents in amperes, not tens of milliamperes. 18 Ω: Would give I = 4.5 / 18 = 0.25 A = 250 mA, far larger than 40 mA. Common Pitfalls: Forgetting to convert mA to A before dividing. Rounding too early and selecting a value off by a factor of 10. Final Answer: 112 Ω

Question 11

Quick current calculation: When 12 V is applied to a 68 Ω resistor, what current flows (choose the closest value)?

Accepted Answer

Introduction / Context: This item reinforces the current form of Ohm’s law, I = V / R, widely used in initial circuit checks and resistor sizing. Correct unit handling ensures that milliampere ranges are interpreted properly. Given Data / Assumptions: Applied voltage V = 12 V. Resistance R = 68 Ω. Ideal linear resistor assumption; temperature and tolerance effects ignored. Concept / Approach: Compute I from I = V / R and convert amperes to milliamperes for comparison with the choices. Retain three significant digits during division, then round to match the nearest listed value. Step-by-Step Solution: Use I = V / R = 12 / 68 A.Compute: 12 / 68 ≈ 0.17647 A.Convert to mA: 0.17647 A ≈ 176 mA. Verification / Alternative check: Back compute voltage: V ≈ I * R ≈ 0.176 * 68 ≈ 11.97 V, which matches the given 12 V within rounding, confirming the selection is correct. Why Other Options Are Wrong: 816 mA: Would require R ≈ 12 / 0.816 ≈ 14.7 Ω, not 68 Ω. 17.6 mA: Ten times too small compared with the correct value. 8.16 mA: Off by roughly a factor of 20. Common Pitfalls: Dropping digits during division and misplacing the decimal when converting to mA. Confusing 68 Ω with 680 Ω due to similar digits. Final Answer: 176 mA

Question 12

Microampere estimation: Approximately how much current flows through a 3.3 MΩ resistor connected across a 30 V source?

Accepted Answer

Introduction / Context: High-value resistors with modest voltages produce microampere-level currents. Estimating these currents is essential for leakage calculations, bias networks, and battery life projections in low-power designs. Given Data / Assumptions: Resistance R = 3.3 MΩ = 3.3 × 10^6 Ω. Voltage V = 30 V. Need current magnitude I in microamperes. Concept / Approach: Ohm’s law gives I = V / R. Converting megaohms to ohms keeps arithmetic consistent; the expected answer is in the microampere range because V is in tens and R is in millions of ohms. Step-by-Step Solution: Compute I = V / R = 30 / (3.3 × 10^6) A.Do the division: 30 / 3.3 ≈ 9.09.Attach the scale: 9.09 × 10^-6 A = 9.09 µA ≈ 9 µA. Verification / Alternative check: Use the handy rule I(µA) ≈ V / R(MΩ). Here 30 / 3.3 ≈ 9.09 µA, which matches the computed value exactly, confirming the microampere estimate. Why Other Options Are Wrong: 90 µA: Would require R ≈ 0.33 MΩ, ten times smaller. 900 µA: Corresponds to R ≈ 33 kΩ, far from 3.3 MΩ. 9000 µA: Implies R ≈ 3.3 kΩ, not megaohms. Common Pitfalls: Confusing MΩ with kΩ and producing milliampere values by mistake. Not converting to microamperes after obtaining the ampere value. Final Answer: 9 µA

Question 13

Voltage requirement by Ohm’s law: What voltage is needed to drive 2.5 A through a 200 Ω resistor (assume steady DC)?

Accepted Answer

Introduction / Context: Determining the required supply voltage for a specified current through a known resistance is a routine design step. It ensures the source and insulation ratings are adequate for the intended load current and power dissipation. Given Data / Assumptions: Resistance R = 200 Ω. Desired current I = 2.5 A. Find voltage V that satisfies Ohm’s law under DC conditions. Concept / Approach: Use the voltage form of Ohm’s law, V = I * R. After computing V, it is wise to evaluate power to judge component stress, since high current through a large resistance implies significant heat. Step-by-Step Solution: Apply V = I * R.Substitute: V = 2.5 * 200.Compute: V = 500 V. Verification / Alternative check: Power check: P = I^2 * R = (2.5)^2 * 200 = 6.25 * 200 = 1250 W. Also P = V * I = 500 * 2.5 = 1250 W. Agreement confirms the calculation and highlights that this scenario demands very high power handling. Why Other Options Are Wrong: 50 V: Would produce I = 0.25 A, ten times smaller. 80 V: Gives I = 0.4 A, well below 2.5 A. 8 V: Gives I = 0.04 A, far too small. Common Pitfalls: Dismissing the power implications; a 200 Ω load at 2.5 A is a kilowatt-class burden. Transposing digits when multiplying, leading to 50 V instead of 500 V. Final Answer: 500 V

Question 14

Convert ohms to milliamperes via Ohm’s law: With a 40 V source and total resistance of 6.8 kΩ, approximately how many milliamperes flow?

Accepted Answer

Introduction / Context: Reading milliampere-level currents from kilohm-level resistances is a standard task in analog design and instrumentation. This problem emphasizes correct prefix conversion and clean division. Given Data / Assumptions: Voltage V = 40 V. Resistance R = 6.8 kΩ = 6800 Ω. Compute current I and express it in mA. Concept / Approach: From Ohm’s law I = V / R. Because R is in kilohms, the result will be in milliamperes; 1 V across 1 kΩ gives 1 mA, a useful mental check known as the 1-k rule of thumb. Step-by-Step Solution: Convert resistance: 6.8 kΩ = 6800 Ω.Compute current: I = 40 / 6800 A ≈ 0.005882 A.Express in mA: 0.005882 A ≈ 5.88 mA ≈ 5.9 mA. Verification / Alternative check: Use the 1-k shortcut: I(mA) ≈ V / R(kΩ) = 40 / 6.8 ≈ 5.88 mA. This quick method matches the detailed computation exactly. Why Other Options Are Wrong: 27.2 mA: Would correspond to about 1.47 kΩ, not 6.8 kΩ. 59 mA: Off by a factor of 10; implies R ≈ 0.68 kΩ. 590 mA: Two orders of magnitude too large for kilohm resistances at tens of volts. Common Pitfalls: Leaving resistance in kΩ while using volts and forgetting the implied milli result. Rounding prematurely and selecting 6 mA vs the more precise 5.9 mA choice. Final Answer: 5.9 mA

Question 15

Set a current limit from a 12 V battery: What resistance value is required to limit current to 3.6 mA?

Accepted Answer

Introduction / Context: Series resistors are commonly used to limit current to safe or specified levels. Selecting the correct ohmic value protects LEDs, sensors, and test circuits powered from batteries or bench supplies. Given Data / Assumptions: Source voltage V = 12 V. Desired current I = 3.6 mA = 0.0036 A. Find R for a simple series circuit: R = V / I. Concept / Approach: Apply Ohm’s law directly after converting mA to A. Expect a value in the kilo-ohm range since milliampere currents from a double-digit voltage typically require thousands of ohms. Step-by-Step Solution: Convert current: 3.6 mA = 0.0036 A.Compute R = V / I = 12 / 0.0036.Calculate: R ≈ 3333.33 Ω ≈ 3.33 kΩ.Choose the closest standard option: 3.3 kΩ. Verification / Alternative check: Back compute the current with 3.3 kΩ: I ≈ 12 / 3300 ≈ 0.00364 A = 3.64 mA, essentially the target 3.6 mA. This validates the selection within rounding and resistor tolerance ranges (for example, 5 percent E24 values). Why Other Options Are Wrong: 33 kΩ and 22 kΩ: Would undercurrent the load severely (tens to hundreds of microamperes). 2.2 kΩ: Would allow I ≈ 5.45 mA, exceeding the limit. Common Pitfalls: Mistaking mA for A and obtaining a result 1000 times too small. Ignoring practical resistor series and tolerances when picking the nearest value. Final Answer: 3.3 kΩ

Question 16

Choose a series resistor: How much resistance is required to draw 17.6 mA from a 12 V source in a simple DC circuit?

Accepted Answer

Introduction / Context: Given a fixed supply voltage and a target current, selecting an appropriate series resistor is a basic design task. This example mirrors LED current setting and bias network sizing in many entry-level circuits. Given Data / Assumptions: Supply voltage V = 12 V. Desired current I = 17.6 mA = 0.0176 A. Simple series circuit; neglect internal source resistance. Concept / Approach: Use R = V / I. Because current is on the order of 10 mA and voltage is about 10 V, the expectation is around hundreds of ohms (V in tens divided by I in hundredths of an ampere). Step-by-Step Solution: Convert current: 17.6 mA = 0.0176 A.Compute R = 12 / 0.0176.Calculate: R ≈ 681.82 Ω ≈ 680 Ω (nearest standard value). Verification / Alternative check: Check current with 680 Ω: I ≈ 12 / 680 ≈ 0.01765 A = 17.65 mA, matching the target within rounding and typical resistor tolerances. Power check: P = I^2 * R ≈ (0.0176)^2 * 680 ≈ 0.21 W; a 0.25 W resistor would be acceptable with modest margin. Why Other Options Are Wrong: 212 Ω: Would draw about 56.6 mA, too high. 6.8 kΩ: Would reduce current to about 1.76 mA, too low. 68 Ω: Would allow about 176 mA, an order of magnitude too large. Common Pitfalls: Selecting 68 Ω or 6.8 kΩ due to similar digits without checking the prefix. Forgetting to evaluate resistor power rating after choosing R. Final Answer: 680 Ω

Question 17

Find resistance from measured current: Twelve volts are applied across an unknown resistor and the measured current is 3 mA. What is the resistor value?

Accepted Answer

Introduction / Context: Translating a measured current into a resistance value is a core diagnostic skill. It is especially helpful when checking unknown components or verifying that a circuit behaves as designed at low currents. Given Data / Assumptions: Applied voltage V = 12 V. Measured current I = 3 mA = 0.003 A. Assume ideal resistor behaviour; DC conditions. Concept / Approach: Rearrange Ohm’s law to R = V / I. Expect a kilo-ohm range because tens of volts divided by milliamps yields thousands of ohms. Meticulous unit conversion prevents large errors. Step-by-Step Solution: Convert current: 3 mA = 0.003 A.Compute resistance: R = 12 / 0.003.Calculate: R = 4000 Ω = 4 kΩ. Verification / Alternative check: Back compute the current using 4 kΩ: I = 12 / 4000 = 0.003 A = 3 mA, which reproduces the measured value exactly. Power check: P = V * I = 12 * 0.003 = 0.036 W, so a 0.125 W or 0.25 W resistor is more than sufficient. Why Other Options Are Wrong: 4 Ω or 4.4 Ω: Would allow multi-ampere currents at 12 V. 400 Ω: Would yield I = 12 / 400 = 30 mA, ten times higher than measured. Common Pitfalls: Dropping the milli prefix in current, leading to 400 Ω instead of 4 kΩ. Confusing kΩ with Ω when reading options quickly. Final Answer: 4 kΩ

Question 18

In a DC circuit powered by an 18 V battery, the ammeter initially reads 40 mA but later drops to 20 mA. Assuming an ideal constant-voltage source, by how many volts has the supply voltage changed, if at all?

Accepted Answer

Introduction / Context: In basic circuit analysis using Ohm's law, it is critical to distinguish between constant-voltage sources and changes in circuit current caused by resistance variations. Here, the source is specified as an 18 V battery, and the question probes whether a change in current necessarily implies a change in supply voltage. Given Data / Assumptions: Supply: ideal 18 V battery (constant-voltage assumption). Initial current: 40 mA. Later current: 20 mA. No explicit change to the source is mentioned; measurement is of circuit current only. Concept / Approach: Ohm's law: V = I * R. With an ideal constant-voltage source, changes in current reflect changes in total circuit resistance or load conditions, not a change in source voltage. Unless the problem states that the battery voltage has drooped or been adjusted, we treat the supply voltage as unchanged. Step-by-Step Solution: Recognize that the source is specified as 18 V (constant).A drop from 40 mA to 20 mA indicates R_total increased (e.g., temperature, contact, or component change).With a constant source, the voltage change ΔV = 0 V. Verification / Alternative check: If the source had changed, the problem would explicitly indicate a new supply value. Since it does not, the standard assumption in DC theory exercises is a fixed source. Current variation alone does not imply voltage variation. Why Other Options Are Wrong: 9 V or 900 mV suggest partial voltage sag without evidence. 18 V implies the supply went to zero, which is not stated. Common Pitfalls: Assuming any current change must come from a changing source; forgetting that resistance variations (or load changes) alter current for a fixed voltage. Final Answer: 0 V

Question 19

Using Ohm’s law for a filament lamp: estimate the filament resistance when the bulb operates from a 110 V source drawing approximately 0.6 A of current.

Accepted Answer

Introduction / Context: Lamp filaments can be modeled as resistive loads for approximate calculations. Ohm’s law provides a quick way to estimate resistance given operating voltage and current. This is useful for sanity checks and rough power calculations in household or lab settings. Given Data / Assumptions: Supply voltage V = 110 V. Operating current I ≈ 0.6 A. Treat the filament as a resistor (temperature effects ignored for this estimate). Concept / Approach: Use the rearranged Ohm’s law: R = V / I. Keep units consistent, compute the ratio, and round sensibly to match the options. Recognize that a real tungsten filament’s resistance varies with temperature, but the question seeks a steady-state approximate value. Step-by-Step Solution: Apply R = V / I.Compute R = 110 / 0.6 ≈ 183.33… Ω.Round to the nearest listed value: 183 Ω. Verification / Alternative check: Power check: P = V * I = 110 * 0.6 = 66 W (a plausible bulb rating), consistent with a filament around a few hundred ohms at operating temperature. Why Other Options Are Wrong: 18.3 Ω or 6.6 Ω would imply extremely high current. 66 Ω would imply I ≈ 1.67 A at 110 V (far larger than stated). Common Pitfalls: Confusing hot and cold resistance; misplacing the decimal when dividing volts by amps. Final Answer: 183 Ω

Question 20

High-voltage DC application: determine the current when 18 kV is applied across a 15 kΩ resistance (state the result in appropriate SI units).

Accepted Answer

Introduction / Context: Ohm’s law scales easily to kilovolt and kilo-ohm ranges common in high-voltage testing and insulation checks. Accurate unit handling (kV, kΩ) prevents order-of-magnitude mistakes and ensures safe design margins. Given Data / Assumptions: Voltage V = 18 kV = 18,000 V. Resistance R = 15 kΩ = 15,000 Ω. Use I = V / R for a purely resistive element. Concept / Approach: Convert both quantities to base SI units (volts, ohms), then compute current. Alternatively, use ratio of prefixes directly: kV/kΩ cancels the kilo factor, leaving amperes. Step-by-Step Solution: Convert: V = 18,000 V, R = 15,000 Ω.Compute I = V / R = 18,000 / 15,000 = 1.2 A.No further scaling is needed; amperes is the correct SI unit for current. Verification / Alternative check: Prefix method: 18 kV / 15 kΩ = (18/15) kV/kΩ = 1.2 A because k cancels. Consistent with the calculation above. Why Other Options Are Wrong: 12 A is 10× too large; 120 mA is 10× too small; 12 mA is 100× too small. Common Pitfalls: Failing to convert kV and kΩ consistently; treating kilo as 1024 instead of 1000 in SI electrical units. Final Answer: 1.2 A

Ohm's Law Questions