AP Physics

Unit 1 - Vectors and Kinematics

MCQ
Mathematical
Intermediate

Pro

Pro

Educator

Upgrade For More Credits
0
Step Reasoning
Identify the target quantity and express the power dissipated by Resistor 1 in terms of the total circuit current \(I\).
\[ P_1 = I^2 R_1 = I^2 R \]
The question asks for a ranking of the power dissipated by each resistor. Expressing all powers in terms of the total current \(I\) entering the parallel network allows a direct comparison.
Use the current divider rule to find the branch currents \(I_2\) and \(I_3\) in terms of total current \(I\).
\[ I_2 = I \left( \dfrac{R_3}{R_2 + R_3} \right) = I \left( \dfrac{6R}{2R + 6R} \right) = \dfrac{3}{4} I \]
\[ I_3 = I \left( \dfrac{R_2}{R_2 + R_3} \right) = I \left( \dfrac{2R}{2R + 6R} \right) = \dfrac{1}{4} I \]
Resistors 2 and 3 are in parallel, so the total current \(I\) splits inversely proportional to their resistance values \(R_2 = 2R\) and \(R_3 = 6R\).
Calculate the power dissipated in Resistors 2 and 3 in terms of \(I^2 R\).
\[ P_2 = I_2^2 R_2 = \left( \dfrac{3}{4} I \right)^2 (2R) = \dfrac{9}{16} I^2 (2R) = \dfrac{9}{8} I^2 R = 1.125 I^2 R \]
\[ P_3 = I_3^2 R_3 = \left( \dfrac{1}{4} I \right)^2 (6R) = \dfrac{1}{16} I^2 (6R) = \dfrac{3}{8} I^2 R = 0.375 I^2 R \]
Knowing the current through each branch allows us to apply \(P = I_{branch}^2 R_{branch}\) and compare directly to \(P_1\).
Compare the power expressions to determine the ranking.
\[ P_2 = 1.125 I^2 R, \quad P_1 = 1.000 I^2 R, \quad P_3 = 0.375 I^2 R \]
\[ \therefore P_2 > P_1 > P_3 \]
Comparing the numerical coefficients of \(I^2 R\) establishes the quantitative ranking from greatest to least.

Why each choice is correct or incorrect:

(A) Incorrectly assumes that because Resistor 1 carries the full circuit current, it must dissipate more power than either parallel resistor, leading to the ranking \(P_1 > P_2 > P_3\).

(B) Incorrectly assumes power dissipation is directly proportional to resistance (\(P \propto R\)) without accounting for current division, resulting in the ranking \(P_3 > P_2 > P_1\).

(C) Incorrectly assumes Resistor 1 dissipates the most power and misapplies current division by assuming the larger resistor in parallel dissipates more power, yielding \(P_1 > P_3 > P_2\).

(D) This is the correct answer. Calculating branch currents yields \(P_2 = 1.125 I^2 R\), \(P_1 = 1.000 I^2 R\), and \(P_3 = 0.375 I^2 R\), giving the ranking \(P_2 > P_1 > P_3\).

Need Help? Ask Phy To Explain

A Major Upgrade To Phy Is Coming Soon — Stay Tuned

Just Drag and Drop!
Quick Actions ?
×

NEW UBQ QUIZ LAB

100s of AP aligned questions and quizzes to help you get a 5 even faster. Full Mock exams with Auto Grading and Adaptive explanations. Try out Nerd Notes', state of the art, quiz platform.

Topics in this question

We'll help clarify entire units in one hour or less — guaranteed.

A self paced course with videos, problems sets, and everything you need to get a 5. Trusted by over 15k students and over 200 schools.

Go Pro to remove ads + unlimited access to our AI learning tools.

D

Nerd Notes

Discover the world's best Physics resources

Continue with

By continuing you (1) agree to our Terms of Use and Terms of Sale and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

Error Report

Sign in before submitting feedback.

KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

Metric Prefixes

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

Sign In to View Your Questions

Share This Question

Enjoying UBQ? Share the 🔗 with friends!

Link Copied!

PRO TIER

One price to unlock most advanced version of Phy across all our tools.

$20

per month

Billed Monthly. Cancel Anytime.

We use cookies to improve your experience. By continuing to browse on Nerd Notes, you accept the use of cookies as outlined in our privacy policy.

Physics Shouldn't Be Hard

So I designed the Ultimate AP Physics 1 Course so you can learn faster and score higher in less than 100 hours.
Trusted by 10k+ Students

📚 Predict Your AP Physics Exam Score

Try our free calculator to see what you need to get a 5 on the 2026 AP Physics 1 exam.

Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed