Physics Ch 21 Circuits

Physics Ch 21 Circuits

10th - 12th Grade

38 Qs

quiz-placeholder

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Physics Ch 21 Circuits

Physics Ch 21 Circuits

Assessment

Quiz

Physics

10th - 12th Grade

Practice Problem

Medium

NGSS
HS-PS2-5, HS-PS3-5, MS-ETS1-1

+1

Standards-aligned

Created by

Kim Fermoyle

Used 70+ times

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38 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Media Image

The symbol shown represents a

battery

resistor

capacitor

transistor

2.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Media Image

The bulbs in the circuit shown are connected __________.

in series

in parallel

to my phone

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Media Image

Which end of the battery shown is the positive terminal?

the end with the longer line (top/ upper end in this image)

the end with the shorter line (bottom/ lower end in this image)

Tags

NGSS.HS-PS2-5

4.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Media Image

The bottom of the light bulb is resting on the positive end of the battery. Does the bulb light?

yes

no

only if it's dark outside

Answer explanation

A circuit has to be a closed path (loop) in order for current to flow through it. Also, the negative end of the battery must be in the circuit.

5.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

The three bulbs are identical and the two batteries are identical. Compare the brightnesses of the bulbs.

A > B > C

A > C > B

A > B = C

A < B = C

A = B = C

6.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

The three bulbs are identical and the two batteries are identical. Compare the brightnesses of the bulbs.

A > B > C

A > C > B

A > B = C

A < B = C

A = B = C

Answer explanation

If you look closely, you can see that the two bulbs are connected in parallel.

7.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Media Image

The potential difference (voltage) across the 10 Ω resistor is

30 V

20 V

15 V

10 V

5 V

Answer explanation

There are two ways to figure this out: 1) Since you know the thirty volts must be used up between the two resistors, and one resistor has twice as much resistance as the other, two-thirds of the voltage must go to the larger resistor. 2) Add the resistances to find the resistance at the battery, use Ohm's Law to find the current in the circuit, then use that current times the resistance to find voltage.

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