What is Electricity & circuits on the Physics Regents?

Electricity & circuits is a Electricity & Magnetism topic on the Physics Regents. RegentsLab has 65 practice questions on it, each with a full worked solution.

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Electricity & circuits: Physics Regents Practice

Practice Electricity & circuits for the Physics Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

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Question 49 · 2 pts

In a metal conductor, 18 coulombs of charge pass a given point in 3.0 seconds. What is the current in the conductor?

(A) 21 A
(B) 0.17 A
(C) 54 A
(D) 6.0 A ✓

Worked solution

$I = \frac{q}{t}$ Current is charge per unit time (reference tables).

$I = \frac{18\ \text{C}}{3.0\ \text{s}}$

$I = 6.0\ \text{A}$

Answer: 6.0 A

Electric current is the rate at which charge flows past a point, so $I=\frac{q}{t}$.

Question 50 · 2 pts

A resistor carries a current of 0.50 A when the potential difference across it is 6.0 V. What is the resistance of the resistor?

(A) 0.083 Ω
(B) 6.5 Ω
(C) 12 Ω ✓
(D) 3.0 Ω

Worked solution

$V = IR \Rightarrow R = \frac{V}{I}$ Ohm's law rearranged for resistance.

$R = \frac{6.0\ \text{V}}{0.50\ \text{A}}$

$R = 12\ \Omega$

Answer: 12 Ω

Ohm's law relates potential difference, current, and resistance; solving for resistance gives $R=\frac{V}{I}$.

Question 58 · 2 pts

Two small charged spheres exert an electrostatic force of magnitude $F$ on each other. If the charges are kept the same but the distance between the spheres is tripled, what is the magnitude of the new electrostatic force?

(A) $9F$
(B) $3F$
(C) $\dfrac{F}{3}$
(D) $\dfrac{F}{9}$ ✓

Worked solution

$F = \frac{k q_1 q_2}{r^2}$ Coulomb's law; force varies as 1/r^2

$r \rightarrow 3r \implies r^2 \rightarrow (3r)^2 = 9r^2$

$F_{new} = \frac{k q_1 q_2}{9r^2} = \frac{1}{9}\,\frac{k q_1 q_2}{r^2} = \frac{F}{9}$

Answer: $F_{new} = \dfrac{F}{9}$

Coulomb's law is an inverse-square law, so tripling the separation reduces the electrostatic force by a factor of $3^2 = 9$.

Question 64 · 2 pts

A resistor dissipates power $P$ when the potential difference across it is $V$. If the potential difference across the same resistor is doubled, how much power does it now dissipate?

(A) $8P$
(B) $4P$ ✓
(C) $P$
(D) $2P$

Worked solution

$P = \frac{V^2}{R}$ Power in terms of voltage (NYS Reference Table); R is fixed

$V \rightarrow 2V \implies P_{new} = \frac{(2V)^2}{R} = \frac{4V^2}{R}$

$P_{new} = 4\,\frac{V^2}{R} = 4P$

Answer: $P_{new} = 4P$

For a fixed resistance, power varies as the square of the voltage ($P = V^2/R$), so doubling the voltage quadruples the dissipated power.

Question 7 · 2 pts

A $12\ \text{V}$ battery is connected across a single $4.0\ \Omega$ resistor. What current flows through the resistor?

(A) $48\ \text{A}$
(B) $0.33\ \text{A}$
(C) $3.0\ \text{A}$ ✓
(D) $4.0\ \text{A}$

Worked solution

$V = IR \Rightarrow I = \dfrac{V}{R}$ Ohm's Law

$I = \dfrac{12\ \text{V}}{4.0\ \Omega}$

$I = 3.0\ \text{A}$

Answer: $3.0\ \text{A}$

By Ohm's Law, current equals voltage divided by resistance; dividing 12 V by 4.0 ohms gives 3.0 A.

Question 19 · 2 pts

A $12\ \Omega$ resistor is connected across a $6.0\text{-V}$ battery. What is the current through the resistor?

(A) $6.5\text{ A}$
(B) $72\text{ A}$
(C) $0.50\text{ A}$ ✓
(D) $2.0\text{ A}$

Worked solution

$I = \dfrac{V}{R}$ Ohm's law

$I = \dfrac{6.0\text{ V}}{12\ \Omega}$

$I = 0.50\text{ A}$

Answer: $0.50\text{ A}$

Ohm's law relates current, voltage, and resistance. Dividing the voltage by the resistance gives a current of 0.50 A.

Question 29 · 2 pts

A 4.0-ohm resistor is connected across the terminals of a 12-volt battery in a simple series circuit. What current flows through the resistor?

(A) 0.33 A
(B) 8.0 A
(C) 48 A
(D) 3.0 A ✓

Worked solution

$V = IR$ Ohm's law

$I = \dfrac{V}{R} = \dfrac{12\ \text{V}}{4.0\ \Omega}$

$I = 3.0\ \text{A}$

Answer: 3.0 A

Ohm's law relates voltage, current, and resistance as V = IR. Solving for current gives I = 12/4.0 = 3.0 A; the 48 A distractor results from multiplying V and R instead of dividing.

Question 46 · 2 pts

A resistor has a potential difference of 12 V across it and a resistance of 4.0 ohms. What is the current through the resistor?

(A) 48 A
(B) 3.0 A ✓
(C) 0.33 A
(D) 8.0 A

Worked solution

$V = IR \Rightarrow I = \dfrac{V}{R}$

$I = \dfrac{12\ \text{V}}{4.0\ \Omega} = 3.0\ \text{A}$

Answer: 3.0 A

Ohm's law states V = IR, so current equals voltage divided by resistance, giving 3.0 A. Inverting the ratio or subtracting the values instead of dividing gives the incorrect distractors.

Question 56 · 4 pts · constructed response

A 12-V battery is connected in series with a $4.0\ \Omega$ resistor and an $8.0\ \Omega$ resistor. Calculate the total resistance of the circuit, the current through the circuit, and the voltage drop across the $8.0\ \Omega$ resistor.

Model answer: Total resistance = 12 Ω; current = 1.0 A; voltage across 8.0 Ω resistor = 8.0 V

Worked solution

$R_{total} = R_1 + R_2 = 4.0 + 8.0 = 12\ \Omega$ series resistances add

$I = \dfrac{V}{R_{total}} = \dfrac{12}{12}$ Ohm's law for the whole circuit

$I = 1.0\ \text{A}$

$V_2 = IR_2 = (1.0)(8.0)$ same current flows through each resistor in series

$V_2 = 8.0\ \text{V}$

Answer: Total resistance = 12 Ω; current = 1.0 A; voltage across 8.0 Ω resistor = 8.0 V

In a series circuit the same current flows through every element, so the total resistance is the simple sum of the individual resistances, and each resistor's voltage drop is found from Ohm's law using that shared current.

Question 63 · 4 pts · constructed response

A 12 V battery is connected in series with a 4.0 $\Omega$ resistor and a 2.0 $\Omega$ resistor. Calculate the current in the circuit and the voltage drop across each resistor.

Model answer: The current is $2.0\ \text{A}$; the voltage drop across the $4.0\ \Omega$ resistor is $8.0\ \text{V}$ and across the $2.0\ \Omega$ resistor is $4.0\ \text{V}$.

Worked solution

$R_{total} = R_1 + R_2 = 4.0\ \Omega + 2.0\ \Omega = 6.0\ \Omega$ series resistances add

$I = \dfrac{V}{R_{total}} = \dfrac{12\ \text{V}}{6.0\ \Omega} = 2.0\ \text{A}$

$V_1 = IR_1 = (2.0\ \text{A})(4.0\ \Omega) = 8.0\ \text{V}, \quad V_2 = IR_2 = (2.0\ \text{A})(2.0\ \Omega) = 4.0\ \text{V}$

Answer: The current is $2.0\ \text{A}$; the voltage drop across the $4.0\ \Omega$ resistor is $8.0\ \text{V}$ and across the $2.0\ \Omega$ resistor is $4.0\ \text{V}$.

In a series circuit the same current flows through every resistor, so the total resistance is the sum of the individual resistances, and each resistor's voltage drop follows from Ohm's law using that shared current.

Question 70 · 4 pts · constructed response

A 12.0-V battery is connected to two resistors, $R_1 = 15.0\ \Omega$ and $R_2 = 25.0\ \Omega$, joined in series. Calculate the current in the circuit and the potential difference across $R_2$.

Model answer: The current in the circuit is 0.300 A, and the potential difference across R2 is 7.50 V.

Worked solution

$R_{eq} = R_1 + R_2 = 15.0\ \Omega + 25.0\ \Omega = 40.0\ \Omega$ series resistors add directly

$I = \dfrac{V}{R_{eq}} = \dfrac{12.0\ \text{V}}{40.0\ \Omega} = 0.300\ \text{A}$ same current flows through all series elements

$V_2 = I R_2 = (0.300\ \text{A})(25.0\ \Omega) = 7.50\ \text{V}$

Answer: The current in the circuit is 0.300 A, and the potential difference across R2 is 7.50 V.

In a series circuit the equivalent resistance is the sum of the individual resistances and the same current flows through every element, so Ohm's law applied to the total circuit gives the current, and applying Ohm's law again to R2 alone gives the voltage drop across it.

Question 8 · 2 pts

A 10. ohm resistor is connected across a 20. V battery. What current flows through the resistor?

(A) 10. A
(B) 0.50 A
(C) 200 A
(D) 2.0 A ✓

Worked solution

$V = IR$ Ohm's law

$I = \dfrac{V}{R} = \dfrac{20.\text{ V}}{10.\ \Omega}$

$I = 2.0\text{ A}$

Answer: 2.0 A

Ohm's law relates voltage, current, and resistance as V = IR, so current equals voltage divided by resistance. Dividing 20. V by 10. ohms gives 2.0 A; multiplying instead of dividing, or inverting the ratio, produces the distractor values.

Question 20 · 2 pts

Two resistors, 6.0 $\Omega$ and 3.0 $\Omega$, are connected in series across a 12-volt battery. What is the current through the 6.0 $\Omega$ resistor?

(A) 1.3 A ✓
(B) 4.0 A
(C) 2.0 A
(D) 0.67 A

Worked solution

$R_{eq} = R_1 + R_2 = 6.0\ \Omega + 3.0\ \Omega = 9.0\ \Omega$

$I = \dfrac{V}{R_{eq}} = \dfrac{12\ \text{V}}{9.0\ \Omega}$

$I = 1.33\ \text{A} \approx 1.3\ \text{A}$

Answer: 1.3 A

In a series circuit the same current flows through every resistor, and that current is found using the total (equivalent) resistance. Using only one resistor's value instead of the sum (giving 2.0 A or 4.0 A) is a common mistake.

Question 31 · 2 pts

A $12\text{-V}$ battery is connected in a series circuit to two resistors of $3.0\ \Omega$ and $9.0\ \Omega$. What is the current in the circuit?

(A) $1.3\ \text{A}$
(B) $1.0\ \text{A}$ ✓
(C) $0.75\ \text{A}$
(D) $4.0\ \text{A}$

Worked solution

$R_{total} = R_1 + R_2 = 3.0 + 9.0 = 12\ \Omega$ Series resistances add

$I = \dfrac{V}{R_{total}} = \dfrac{12\ \text{V}}{12\ \Omega}$

$I = 1.0\ \text{A}$

Answer: $1.0\ \text{A}$

In a series circuit the same current flows through all components; the total resistance is the sum of the individual resistances, and Ohm's law gives the current.

Question 33 · 2 pts

A $10\ \Omega$ resistor carries a current of $2.0\ \text{A}$. How much power is dissipated by the resistor?

(A) $40\ \text{W}$ ✓
(B) $20\ \text{W}$
(C) $10\ \text{W}$
(D) $200\ \text{W}$

Worked solution

$P = I^2 R$ Power dissipated by a resistor

$P = (2.0)^2 (10) = (4.0)(10)$

$P = 40\ \text{W}$

Answer: $40\ \text{W}$

Power dissipated in a resistor can be found from $P=I^2R$; a common error is to use $P=IR$, which is dimensionally incorrect and gives voltage, not power.

Question 44 · 2 pts

A 3.0-ohm resistor and a 9.0-ohm resistor are connected in series to a 12-volt battery. What is the current in the circuit?

(A) 0.25 A
(B) 4.0 A
(C) 1.3 A
(D) 1.0 A ✓

Worked solution

$R_{total} = R_1 + R_2 = 3.0\ \Omega + 9.0\ \Omega = 12\ \Omega$

$I = \dfrac{V}{R_{total}} = \dfrac{12\text{ V}}{12\ \Omega} = 1.0\text{ A}$

Answer: 1.0 A

In a series circuit the resistances add directly; dividing the battery voltage by the total series resistance of 12 ohms gives a current of 1.0 A.

Question 50 · 2 pts

A 4.0-ohm resistor and an 8.0-ohm resistor are connected in series to a 24-volt battery. What is the current through the 4.0-ohm resistor?

(A) 9.0 A
(B) 3.0 A
(C) 6.0 A
(D) 2.0 A ✓

Worked solution

$R_{eq} = R_1 + R_2 = 4.0\ \Omega + 8.0\ \Omega = 12\ \Omega$ Series resistors add directly

$I = \dfrac{V}{R_{eq}} = \dfrac{24\text{ V}}{12\ \Omega} = 2.0\text{ A}$ Ohm's law; same current through all series elements

Answer: 2.0 A

In a series circuit the same current flows through every resistor, and the total resistance is the sum of the individual resistances. Dividing the battery voltage by the equivalent series resistance gives the current. A student who divides 24 V by only the 4.0-ohm resistor (ignoring the series combination) incorrectly gets 6.0 A, and one who mistakenly treats the resistors as parallel gets 9.0 A.

Question 56 · 4 pts · constructed response

A 4.0 ohm resistor and an 8.0 ohm resistor are connected in series with a 12 V battery. Calculate the current in the circuit and the voltage drop across the 8.0 ohm resistor.

Model answer: The current in the circuit is 1.0 A, and the voltage drop across the 8.0 ohm resistor is 8.0 V.

Worked solution

$R_{eq} = R_1 + R_2 = 4.0 + 8.0 = 12\ \Omega$ Resistors in series add directly

$I = \dfrac{V}{R_{eq}} = \dfrac{12}{12} = 1.0\ \text{A}$ Same current flows through every element in a series circuit

$V_{8.0} = IR_{8.0} = (1.0)(8.0) = 8.0\ \text{V}$

Answer: The current in the circuit is 1.0 A, and the voltage drop across the 8.0 ohm resistor is 8.0 V.

In a series circuit the total resistance is the sum of individual resistances and the same current flows through each resistor, so Ohm's law applied to the 8.0 ohm resistor gives its voltage drop.

Question 8 · 2 pts

A resistor carries a current of $2.0\ \text{A}$ and has a resistance of $5.0\ \Omega$. What is the potential difference across the resistor?

(A) $2.5\ \text{V}$
(B) $7.0\ \text{V}$
(C) $10.\ \text{V}$ ✓
(D) $0.40\ \text{V}$

Worked solution

$V = IR$ Ohm's law

$V = (2.0\ \text{A})(5.0\ \Omega)$

$V = 10.\ \text{V}$

Answer: $10.\ \text{V}$

Ohm's law relates voltage, current, and resistance; multiplying the given current and resistance gives the potential difference of 10 V.

Question 22 · 2 pts

A resistor with resistance $4.0\ \Omega$ is connected across a $12\ \text{V}$ battery. What is the current through the resistor?

(A) $48\ \text{A}$
(B) $0.33\ \text{A}$
(C) $16\ \text{A}$
(D) $3.0\ \text{A}$ ✓

Worked solution

$V = IR \;\Rightarrow\; I = \dfrac{V}{R}$ Ohm's law

$I = \dfrac{12\ \text{V}}{4.0\ \Omega}$

$I = 3.0\ \text{A}$

Answer: $3.0\ \text{A}$

Ohm's law relates voltage, current, and resistance; dividing the voltage by the resistance gives the current through the resistor.

Question 47 · 2 pts

A 12 ohm resistor is connected across a 6.0 V battery in a simple series circuit. What is the current through the resistor?

(A) 18 A
(B) 0.50 A ✓
(C) 72 A
(D) 2.0 A

Worked solution

$V = IR \;\Rightarrow\; I = \dfrac{V}{R}$ Ohm's law

$I = \dfrac{6.0\ \text{V}}{12\ \Omega} = 0.50\ \text{A}$

Answer: 0.50 A

Ohm's law relates voltage, current, and resistance as $V=IR$. Solving for current gives $I = V/R = 6.0/12 = 0.50$ A. Inverting the ratio (12/6) gives the distractor 2.0 A.

Question 56 · 4 pts · constructed response

A 6.00 Ω resistor and a 12.0 Ω resistor are connected in series with an 18.0 V battery. Calculate the current in the circuit and the voltage drop across the 12.0 Ω resistor.

Model answer: The current is $1.00\ \text{A}$ and the voltage across the $12.0\ \Omega$ resistor is $12.0\ \text{V}$.

Worked solution

$R_{eq} = R_1 + R_2 = 6.00 + 12.0 = 18.0\ \Omega$ Series resistors add directly

$I = \dfrac{V}{R_{eq}} = \dfrac{18.0}{18.0} = 1.00\ \text{A}$

$V_{12} = IR_{12} = (1.00)(12.0) = 12.0\ \text{V}$

Answer: The current is $1.00\ \text{A}$ and the voltage across the $12.0\ \Omega$ resistor is $12.0\ \text{V}$.

In a series circuit the current is the same through every resistor, so the total current is found from the total resistance and source voltage, and the voltage across any one resistor is that current times its own resistance.

Question 4 · 2 pts

A 12 V battery is connected in series with a 2.0 Ω resistor and a 4.0 Ω resistor. What is the current in the circuit?

(A) 3.0 A
(B) 6.0 A
(C) 2.0 A ✓
(D) 0.5 A

Worked solution

$R_{total} = 2.0 + 4.0 = 6.0\ \Omega$ series resistors add directly

$I = V/R_{total} = 12/6.0 = 2.0\ A$

Answer: 2.0 A

In a series circuit the resistances add, and the total current is found from Ohm's law using the equivalent resistance.

Question 33 · 2 pts

A resistor with resistance $4.0\ \Omega$ has a potential difference of $12\ \text{V}$ across it. What is the current through the resistor?

(A) 8.0 A
(B) 48 A
(C) 0.33 A
(D) 3.0 A ✓

Worked solution

$I = \dfrac{V}{R}$

$I = \dfrac{12\ \text{V}}{4.0\ \Omega}$

$I = 3.0\ \text{A}$

Answer: 3.0 A

Ohm's law gives current as voltage divided by resistance, $I = V/R = 12/4.0 = 3.0$ A.

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