Physics 2 Quiz: Using Ammeters And Voltmeters

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Using Ammeters And VoltmetersQuestion 1 of 7

A student has a circuit with a 9 V battery (assumed ideal) and three resistors in a configuration where $R_1 = 100\ \Omega$ is in series with the parallel combination of $R_2 = 200\ \Omega$ and $R_3 = 200\ \Omega$ . An ammeter with internal resistance $R_A = 2\ \Omega$ is inserted in series with $R_2$ only (not in the main line and not in series with $R_3$ ).

Compared to the ideal case (zero-resistance ammeter), inserting this real ammeter changes the current through $R_3$ . Which of the following correctly describes the direction and approximate magnitude of this change in $R_3$ 's current?

The current through

R_3

decreases by approximately 0.5 mA, because the ammeter adds resistance to the

R_2

branch, reducing the total current drawn from the battery, and this reduction in total current lowers the current available to all parallel branches including

R_3

The current through

R_3

increases by approximately 0.9 mA, because inserting the ammeter in the

R_2

branch raises that branch's resistance significantly, diverting a large fraction of the parallel-branch current into the

R_3

branch.

The current through

R_3

increases by approximately 0.06 mA, because the ammeter resistance in the

R_2

branch raises its impedance, redirecting a small fraction of the parallel-branch current to

R_3

, while the total voltage across the parallel combination changes only slightly.

The current through

R_3

decreases by approximately 0.06 mA, because adding resistance to the

R_2

branch lowers the total conductance of the parallel combination, which raises the voltage drop across

R_1

and reduces the voltage available across the parallel section, lowering current through both

R_2

and

R_3

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Physics 2 Quiz

Physics 2 Quiz: Using Ammeters And Voltmeters

Practice Using Ammeters And Voltmeters in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Using Ammeters And Voltmeters, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The current through $R_3$ decreases by approximately 0.5 mA, because the ammeter adds resistance to the $R_2$ branch, reducing the total current drawn from the battery, and this reduction in total current lowers the current available to all parallel branches including $R_3$ .
The current through $R_3$ increases by approximately 0.9 mA, because inserting the ammeter in the $R_2$ branch raises that branch's resistance significantly, diverting a large fraction of the parallel-branch current into the $R_3$ branch.
The current through $R_3$ increases by approximately 0.06 mA, because the ammeter resistance in the $R_2$ branch raises its impedance, redirecting a small fraction of the parallel-branch current to $R_3$ , while the total voltage across the parallel combination changes only slightly. (correct answer)
The current through $R_3$ decreases by approximately 0.06 mA, because adding resistance to the $R_2$ branch lowers the total conductance of the parallel combination, which raises the voltage drop across $R_1$ and reduces the voltage available across the parallel section, lowering current through both $R_2$ and $R_3$ .

Explanation: When a real ammeter is inserted into one branch of a parallel combination, two competing effects occur simultaneously: the added resistance makes that branch harder for current to flow through (pushing some current toward the other branch), but it also slightly reduces the total current from the battery (which would pull current away from all branches). The key insight is figuring out which effect dominates — and by how much. Start with the ideal case. The parallel combination of

R_2 = 200\ \Omega

and

R_3 = 200\ \Omega

gives

R_{parallel} = 100\ \Omega

, so total resistance is

100 + 100 = 200\ \Omega

. Battery current is

9/200 = 45\ \text{mA}

, and the voltage across the parallel section is

45 \times 100 = 4.5\ \text{V}

. Current through

R_3

4.5/200 = 22.5\ \text{mA}

. Now insert the ammeter:

R_2

's branch becomes

202\ \Omega

. New parallel resistance:

\frac{200 \times 202}{200 + 202} \approx 100.5\ \Omega

. Total resistance

\approx 200.5\ \Omega

, battery current

\approx 44.89\ \text{mA}

, and parallel voltage

\approx 44.89 \times 100.5 \approx 4.511\ \text{V}

. Current through

R_3

4.511/200 \approx 22.56\ \text{mA}

. That's an increase of about 0.06 mA — confirming C. A is wrong because it claims

R_3

's current decreases, but the redistribution effect (current fleeing the higher-resistance

R_2

branch) outweighs the small reduction in total battery current. B correctly identifies the direction (increase) but wildly overestimates the magnitude — a

2\ \Omega

addition to a

200\ \Omega

branch is tiny, not dramatic. D mirrors A's directional error with a plausible-sounding but incorrect mechanism. Your takeaway: when a small resistance is added to one parallel branch, always calculate both effects numerically rather than relying on intuition — the dominant effect isn't always obvious, and the magnitudes matter just as much as the direction.

Question 2

An ideal voltmeter and an ideal ammeter are available. A student wants to measure the resistance of a single unknown resistor R using the ammeter-voltmeter method. She can either (1) connect the voltmeter directly across R and place the ammeter between the voltage source and the voltmeter–R parallel combination, or (2) connect the ammeter in series with R and place the voltmeter across both the ammeter and R together. Both meters are ideal. Which statement correctly compares the two configurations when the meters are ideal?

Configuration 1 gives a more accurate result because the voltmeter, being ideal, draws no current, so the ammeter measures only the current through R, while in Configuration 2 the voltmeter reads a voltage that includes the ammeter's drop.
Both configurations give exactly the same, perfectly accurate result for R because ideal meters introduce no measurement error by definition, regardless of their placement in the circuit. (correct answer)
Configuration 2 gives a more accurate result because placing the voltmeter outside the ammeter prevents the voltmeter from loading the source, whereas in Configuration 1 the ammeter intercepts current that bypasses R through the voltmeter branch.
Configuration 1 gives a more accurate result because the ammeter is placed closer to the voltage source, reducing the effect of lead resistance, while Configuration 2 introduces systematic error due to the voltmeter's proximity to ground potential.

Explanation: Whenever you see a question involving the ammeter-voltmeter method, your first instinct should be to ask: "What error does each meter introduce?" The key here is the word ideal. An ideal voltmeter has infinite resistance (draws zero current), and an ideal ammeter has zero resistance (drops zero voltage). With those definitions in mind, both configurations become perfectly accurate. In Configuration 1, the ideal voltmeter draws no current, so the ammeter reads exactly the current through R, and the voltmeter reads exactly the voltage across R — giving

R = V/I

with no error. In Configuration 2, the ideal ammeter has zero resistance, so the voltmeter reads only the voltage across R (not across the ammeter), and the current through R is exactly what the ammeter measures — again giving a perfect result. Since neither meter distorts the circuit in any way, both configurations yield the exact same, perfectly accurate value of R. That makes B correct. Answer A describes real-world reasoning accurately — a real voltmeter with finite resistance would cause issues in Configuration 2 — but the question specifies ideal meters, so this concern evaporates entirely. Answer C makes the opposite real-world argument (favoring Configuration 2), but again, the "loading" problem only exists for non-ideal meters. Answer D introduces a completely fabricated concept — "proximity to ground potential" and "lead resistance" are not relevant systematic errors in this idealized scenario, making D a distractor built on plausible-sounding but nonsensical physics. Your takeaway: whenever a problem says "ideal meters," treat voltmeters as open circuits and ammeters as short circuits. All configuration-dependent errors disappear, and meter placement becomes irrelevant to accuracy.

Question 3

In a Wheatstone bridge circuit, four resistors $P$ , $Q$ , $R$ , and $S$ are arranged in the standard diamond configuration with a battery across one diagonal. A galvanometer is connected across the other diagonal to detect balance. At balance, the galvanometer reads zero. A student argues: 'Since the galvanometer reads zero at balance, it doesn't matter whether the galvanometer has a resistance of 1 Ω or 1 MΩ — the balance condition and the values of the unknown resistor determined from $P/Q = R/S$ are completely unaffected by the galvanometer's resistance.' Which of the following most accurately evaluates this claim?

The claim is entirely correct: at balance, no current flows through the galvanometer regardless of its resistance, so the galvanometer resistance has no effect on the balance condition, the balance point itself, or the sensitivity with which the bridge detects small departures from balance.
The claim is partially correct: the balance condition $P/Q = R/S$ and the balance point are indeed independent of galvanometer resistance, but the sensitivity of the bridge — its ability to detect small deviations from balance — does depend on galvanometer resistance, so a high-resistance galvanometer reduces the bridge's ability to precisely locate the balance point. (correct answer)
The claim is incorrect: the balance condition $P/Q = R/S$ is only an approximation that assumes an ideal (zero-resistance) galvanometer; a real galvanometer with finite resistance shifts the true balance point and causes a systematic error in the determined unknown resistance.
The claim is partially correct: the balance condition $P/Q = R/S$ holds only when the galvanometer resistance equals the geometric mean $\sqrt{PS}$ of the bridge arms; for all other galvanometer resistances, the balance condition must be corrected by a factor that accounts for the galvanometer's loading of the midpoint potentials.

Explanation: Whenever you encounter a Wheatstone bridge problem, separate two distinct ideas: the balance condition (where equilibrium occurs) and the sensitivity (how easily you can detect departures from that equilibrium). Students often conflate these, which is exactly the trap this question sets. The balance condition

P/Q = R/S

is derived by requiring that both midpoints of the bridge sit at identical potentials, so zero current flows through the galvanometer branch. Crucially, if zero current flows through that branch, the galvanometer's resistance creates zero voltage drop — its value is simply irrelevant to where balance occurs. You can verify this with Kirchhoff's laws: the balance condition emerges purely from the four arm resistors, with the galvanometer branch dropping out of the equations entirely. So the student is right that

P/Q = R/S

is unaffected by galvanometer resistance. However, sensitivity is a different story. When the bridge is slightly off balance, a small current flows through the galvanometer. A high-resistance galvanometer suppresses that deflection current, making it harder to detect the departure from balance and harder to precisely locate the null point. This is why sensitive laboratory bridges use low-resistance galvanometers. The student's claim ignores this, making it only partially correct — which confirms B as the best answer. A is wrong because it incorrectly extends the argument to sensitivity, which does depend on galvanometer resistance. C is wrong because the balance condition is exact, not an approximation requiring an ideal galvanometer — no systematic error in

P/Q = R/S

arises from finite galvanometer resistance. D introduces a fictional correction factor involving

\sqrt{PS}

that has no basis in the actual circuit analysis. Study tip: On bridge circuit questions, always ask two separate questions — "where is balance?" and "how detectable is imbalance?" They have different answers and depend on different variables.

Question 4

A student constructs a circuit consisting of a real battery (EMF = 12 V, internal resistance r = 2 Ω) connected to two resistors in series: R₁ = 8 Ω and R₂ = 14 Ω. The student wishes to measure the terminal voltage of the battery and the current through R₂ simultaneously. She inserts an ammeter (internal resistance = 0.5 Ω) in series with R₂ and connects a voltmeter (internal resistance = 10 kΩ) across the battery terminals.

Which of the following best describes how the ammeter and voltmeter readings differ from the ideal values that would exist if both meters were ideal (zero-resistance ammeter, infinite-resistance voltmeter)?

The ammeter reads slightly lower than the ideal current because adding its resistance reduces the total circuit current, and the voltmeter reads slightly lower than the ideal terminal voltage because the voltmeter draws a small additional current that increases the voltage drop across the internal resistance. (correct answer)
The ammeter reads slightly lower than the ideal current because adding its resistance reduces the total circuit current, and the voltmeter reads slightly higher than the ideal terminal voltage because the voltmeter's finite resistance creates a parallel path that effectively raises the measured voltage.
The ammeter reads the same as the ideal current because its resistance is negligible compared to the total circuit resistance, and the voltmeter reads slightly lower than the ideal terminal voltage because the voltmeter draws a small additional current that increases the voltage drop across the internal resistance.
The ammeter reads slightly higher than the ideal current because the voltmeter draws additional current that flows through the ammeter branch, and the voltmeter reads slightly higher than the ideal terminal voltage because the parallel combination of R₁ and the voltmeter reduces the effective load resistance.

Explanation: Whenever a question involves real meters in a circuit, your job is to ask: how does each meter's non-ideal resistance perturb the circuit? An ideal ammeter has zero resistance; a real one adds resistance in series. An ideal voltmeter has infinite resistance; a real one draws a small extra current. Here's the reasoning that confirms A is correct. The ideal total resistance is

r + R_1 + R_2 = 2 + 8 + 14 = 24 \, \Omega

, giving an ideal current of

I_{ideal} = 12/24 = 0.5 \, \text{A}

. The real ammeter (0.5 Ω) adds to the series loop, making the total resistance 24.5 Ω and reducing the current to roughly 0.490 A — slightly lower. For the voltmeter: even though 10 kΩ is large, it pulls a tiny extra current (about 1.17 mA) from the battery. That extra current increases the drop across the internal resistance, which lowers the terminal voltage. Both effects match choice A exactly. Choice B is wrong because a finite-resistance voltmeter doesn't raise the terminal voltage — drawing more current from the battery always increases the internal voltage drop, reducing terminal voltage. Choice C is tempting because 0.5 Ω seems negligible, but the question asks whether readings differ from ideal — any nonzero resistance does cause a measurable (if small) reduction, so "reads the same" is physically incorrect. Choice D is wrong on both counts: the voltmeter is across the battery terminals, not in parallel with R₁, and drawing extra current lowers — not raises — the terminal voltage. Your study tip: always trace where each meter sits in the circuit. Series placement (ammeter) increases total resistance; parallel placement (voltmeter) increases total current drawn — both effects work against the quantity being measured.

Question 5

Two identical resistors, each with resistance $R = 1\ \text{k}\Omega$ , are connected in series across an ideal 10 V DC source. A student uses a digital voltmeter with input impedance $R_V = 10\ \text{M}\Omega$ to measure the voltage across one of the resistors.

The student then replaces the digital voltmeter with an analog voltmeter on its 10 V range, which has a sensitivity of $20\ \text{k}\Omega/\text{V}$ (so its internal resistance on the 10 V range is 200 kΩ). Both voltmeters are connected across the same resistor. By approximately how much does the analog voltmeter reading differ from the digital voltmeter reading, and in which direction?

The analog voltmeter reads approximately 0.24 V lower than the digital voltmeter, because the analog voltmeter's 200 kΩ resistance in parallel with the 1 kΩ resistor significantly reduces the effective resistance at those nodes, pulling the measured voltage well below the ~5.0 V that the digital meter reads.
The analog voltmeter reads approximately 0.012 V higher than the digital voltmeter, because the lower input impedance of the analog meter causes more current to flow from the source, increasing the voltage drop across the measured resistor relative to the other resistor in the series chain.
The analog voltmeter reads approximately 0.025 V lower than the digital voltmeter, because both meters draw current from the source and the analog meter's lower impedance increases the total current, raising the voltage drop across the unmeasured resistor and reducing the reading across the measured one.
The analog voltmeter reads approximately 0.012 V lower than the digital voltmeter, because the analog meter's 200 kΩ resistance slightly loads the resistor by creating a parallel combination of ~0.995 kΩ, shifting the voltage divider ratio by a small but nonzero amount from the near-ideal 5.000 V reading. (correct answer)

Explanation: Whenever a voltmeter is placed across a circuit element, it forms a parallel combination with that element — this is called voltmeter loading, and it always pulls the measured voltage slightly downward. The key question is how much. Start with the digital voltmeter. Its

R_V = 10\ \text{M}\Omega

in parallel with

R = 1\ \text{k}\Omega

gives an effective resistance of essentially

1\ \text{k}\Omega

(the 10 MΩ barely matters), so the digital meter reads almost exactly

5.000\ \text{V}

. Now use the analog meter. Its

200\ \text{k}\Omega

in parallel with

1\ \text{k}\Omega

gives:

R_{\text{eff}} = \frac{1000 \times 200000}{1000 + 200000} \approx 995\ \Omega

The voltage divider is now

1000\ \Omega

(unmeasured resistor) versus

995\ \Omega

(measured side), so the analog reading is:

V_{\text{analog}} = 10 \times \frac{995}{1995} \approx 4.988\ \text{V}

The difference is

5.000 - 4.988 \approx 0.012\ \text{V}

lower — confirming D. A is wrong because it wildly overestimates the loading effect, claiming a 0.24 V drop; the parallel combination barely shifts the resistance. B is wrong in both magnitude and direction — a meter in parallel reduces effective resistance and lowers the reading, never raises it. C identifies the correct direction but reports 0.025 V, which corresponds to a calculation error in the voltage divider. As a study tip: when evaluating voltmeter loading, always compute the parallel combination first, then re-run the voltage divider — a high-impedance meter barely shifts the ratio, while a low-impedance one can noticeably distort it.

Question 6

A galvanometer has a full-scale deflection current of $I_g = 500\ \mu$ A and a coil resistance of $R_g = 100\ \Omega$ . An engineer converts this galvanometer into a multi-range ammeter by adding shunt resistors. For Range 1, a shunt $S_1$ is added to allow full-scale reading at $I_1 = 50$ mA. For Range 2, a different shunt $S_2$ is added (replacing $S_1$ , not in addition to it) to allow full-scale reading at $I_2 = 500$ mA.

The engineer mistakenly connects the meter on Range 2 (designed for 500 mA full scale) into a circuit where the actual current is 45 mA, and reads the deflection as approximately 9/100 of full scale. She then switches the range selector to Range 1 (designed for 50 mA full scale) without disconnecting the meter from the circuit first. Assuming the circuit maintains a constant current of 45 mA, what happens immediately after switching to Range 1?

The galvanometer deflects to approximately 9/10 of full scale on Range 1, correctly showing 45 mA, because switching to Range 1 places the lower shunt $S_1$ across the galvanometer and the 45 mA now properly distributes between the galvanometer and $S_1$ for a near-full-scale reading.
The galvanometer deflects to approximately 9/10 of full scale on Range 1, correctly showing 45 mA in steady state, but the galvanometer coil is briefly exposed to excessive current during the switching transient because $S_2$ is momentarily disconnected before $S_1$ is connected (break-before-make switching), potentially damaging the coil. (correct answer)
The galvanometer reading is unaffected during switching because the total circuit resistance changes negligibly: $S_1$ has a higher resistance than $S_2$ , so replacing $S_2$ with $S_1$ increases the total shunt resistance only slightly, and the galvanometer current remains within safe limits throughout the transition.
The galvanometer immediately pegs beyond full scale and may be damaged, because during the switching transient the shunt path is momentarily open, forcing all 45 mA — far exceeding the 0.5 mA rated full-scale current — to flow directly through the galvanometer coil.

Explanation: When working with galvanometers and shunt-based ammeters, always think about what happens to the current path during a range switch, not just before and after. Here's the key physics: a galvanometer is extremely sensitive — its full-scale current is only

I_g = 500\ \mu\text{A}

. Shunt resistors work by providing a low-resistance bypass path, so most of the measured current flows through the shunt rather than the delicate coil. The galvanometer only ever "sees" a tiny fraction of the total current. This protection depends entirely on the shunt being connected. If the range selector uses break-before-make switching — meaning it disconnects the old shunt (

S_2

) before connecting the new one (

S_1

) — there's a brief moment where no shunt exists. During that transient, all 45 mA is forced through the coil. Since full-scale is only 0.5 mA, this represents 90× overload, which can burn out the coil wire or permanently magnetize the core. Once switching completes and

S_1

is in place, the steady-state reading correctly settles at 9/10 of full scale. This makes B correct: the final reading is accurate, but a dangerous transient occurs in between. A is wrong because it ignores the switching transient entirely — real mechanical range selectors don't teleport from one shunt to the next instantaneously. C is wrong on two counts:

S_1

actually has a much higher resistance than

S_2

(lower range = higher shunt resistance), and this irrelevant detail doesn't address the open-circuit moment. D is wrong because it claims the meter "immediately pegs" as a permanent result, ignoring that once

S_1

connects, the reading normalizes — the damage risk is transient, not sustained. Strategy tip: On questions involving switching or disconnecting measuring instruments, always ask yourself: what happens to every current path during the transition? An unprotected galvanometer, even for milliseconds, can be destroyed by overcurrent.

Question 7

A student wants to measure the current through and voltage across a light-emitting diode (LED) operating in forward bias, where the LED's dynamic resistance varies nonlinearly with current. She has an ammeter ( $R_A = 5\ \Omega$ ) and a voltmeter ( $R_V = 10\ \text{k}\Omega$ ). She must choose between: Method I — voltmeter directly across the LED, ammeter in series with the LED-voltmeter parallel combination (ammeter external); Method II — ammeter in series with the LED, voltmeter across the LED-ammeter series combination (voltmeter external). She calculates $R_{LED} = V_{measured}/I_{measured}$ for each method. Which statement correctly identifies the systematic bias introduced by each method?

Method I always overestimates $R_{LED}$ because the ammeter reads the sum of the LED current and voltmeter current, making $I_{measured} > I_{LED}$ , while $V_{measured} = V_{LED}$ correctly; Method II always underestimates $R_{LED}$ because the voltmeter reads the sum of the LED voltage drop and the ammeter voltage drop, making $V_{measured} > V_{LED}$ , while $I_{measured} = I_{LED}$ correctly.
Both methods introduce the same magnitude of error but in opposite directions, so averaging the two calculated values of $R_{LED}$ from both methods yields the true dynamic resistance of the LED with no systematic bias, regardless of the LED's operating point.
Method I underestimates $R_{LED}$ only when $R_{LED} > \sqrt{R_A \cdot R_V}$ , and overestimates it otherwise; Method II overestimates $R_{LED}$ only when $R_{LED} < \sqrt{R_A \cdot R_V}$ , and the two methods give equal accuracy at $R_{LED} = \sqrt{R_A \cdot R_V}$ .
Method I always underestimates $R_{LED}$ because the ammeter reads the sum of the LED current and voltmeter current, making $I_{measured} > I_{LED}$ , while $V_{measured} = V_{LED}$ correctly; Method II always overestimates $R_{LED}$ because the voltmeter reads the sum of the LED voltage drop and the ammeter voltage drop, making $V_{measured} > V_{LED}$ , while $I_{measured} = I_{LED}$ correctly. (correct answer)

Explanation: When measuring resistance indirectly using an ammeter and voltmeter, you must always ask: what does each instrument actually read? The key insight is that real meters have non-ideal resistances that corrupt one of your two measurements. In Method I (ammeter external), the voltmeter sits directly across the LED, so

V_{measured} = V_{LED}

exactly. However, the ammeter is in series with the parallel combination of the LED and voltmeter, so it reads the total current:

I_{measured} = I_{LED} + I_V

. Since

I_{measured} > I_{LED}

, the calculated

R_{LED} = V_{measured}/I_{measured}

is smaller than the true value — Method I underestimates $R_{LED}$ . In Method II (voltmeter external), the ammeter is in series with the LED only, so

I_{measured} = I_{LED}

exactly. But the voltmeter spans both the LED and ammeter, giving

V_{measured} = V_{LED} + V_A = V_{LED} + I_{LED} \cdot R_A

. Since

V_{measured} > V_{LED}

, the calculated resistance is larger than the true value — Method II overestimates $R_{LED}$ . This confirms D. A reverses the bias direction for both methods — a classic trap if you mix up which instrument is corrupted in each configuration. B is false: the errors are not equal in magnitude, so averaging does not cancel systematic bias. C introduces a crossover condition (

\sqrt{R_A \cdot R_V}

) that applies to choosing the better method, not to the direction of bias — both methods have fixed, unconditional bias directions. Study tip: Memorize the rule — ammeter external →

I

is too high → underestimate

R

; voltmeter external →

V

is too high → overestimate

R

. These directions never flip.