Physics: Circuits

The symbol shown represents a

• battery

• resistor

• capacitor

• transistor

The bulbs in the circuit shown are connected __________.

• in series

• in parallel

• to my phone

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.

Are bulbs B and C connected in series or in parallel?

• There is no way to tell

• Parallel

• Series

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

Are bulbs B and C connected in series or in parallel?

• There is no way to tell

• Parallel

• Series

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.

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.

What thing(s) about the resistors in this circuit is/are the same for all three?

• Current I

• Potential difference ΔV

• Resistance R

• A and B

Answer explanation

Voltage is always the same for devices connected in parallel.

The battery current I is

• 3 A

• 2 A

• 1 A

• 2/3 A

• 1/2 A

Answer explanation

To solve: add the resistors to find the resistance at the battery, then use Ohm's Law.