Quiz · Medium
Ohm's law, series and parallel resistors, power, energy cost, Kirchhoff's rules and capacitors. Numeric practice for high school and intro college physics.
At a junction, 5.0 A flows in. One branch carries 2.0 A out. How much current flows in the other branch?
3.0 A
7.0 A
2.5 A
5.0 A
Kirchhoff's junction rule: current in = current out, because charge is conserved. 5.0 − 2.0 = 3.0 A.
A 12 V battery is connected across a 4.0 Ω resistor. What current flows?
48 A
0.33 A
16 A
3.0 A
Ohm's law: . 48 multiplies V and R, and 0.33 divides R by V.
Resistors of 2 Ω, 3 Ω and 5 Ω are connected in series. What is the total resistance?
10 Ω
About 0.97 Ω
30 Ω
5 Ω
In series, resistances add: . About 0.97 Ω is what the parallel formula gives.
A 6 Ω resistor and a 3 Ω resistor are connected in parallel. What is the equivalent resistance?
9 Ω
2 Ω
4.5 Ω
18 Ω
, so . For two resistors you can also use product over sum: 18 / 9 = 2 Ω. The parallel total is always less than the smallest resistor.
Two identical 10 Ω resistors are connected in parallel. What is the equivalent resistance?
20 Ω
10 Ω
0.2 Ω
5 Ω
For n identical resistors in parallel, . 0.2 is , a common place to stop too early.
A 2 Ω and a 4 Ω resistor are in series with a 12 V battery. What is the voltage across the 4 Ω resistor?
4 V
8 V
12 V
6 V
Total R = 6 Ω, so . Voltage across the 4 Ω resistor = (and 4 V across the 2 Ω, adding to 12 V). In series, the bigger resistor takes the bigger share.
A device on a 120 V outlet draws 0.50 A. What power does it use?
60 W
240 W
0.0042 W
120 W
. 240 W divides V by I, which gives resistance in ohms, not power.
What power is dissipated when 9.0 V is applied across a 3.0 Ω resistor?
3.0 W
81 W
9.0 W
27 W
. Check: and . 3.0 is the current, not the power.
A 100 W bulb runs for 10 hours. Electricity costs $0.15 per kWh. How much does it cost?
$1.50
$0.15
$150
$0.015
Energy = 0.100 kW × 10 h = 1.0 kWh. Cost = 1.0 × $0.15 = $0.15. Forgetting to convert watts to kilowatts gives $150.
Three bulbs are wired in parallel. One burns out. What happens to the other two?
They all go out
They get brighter and burn out
They flicker on and off
They stay lit
Each parallel branch has its own complete path to the battery, so the others keep working at the same brightness with an ideal battery. In a series circuit, one break opens the only path and every bulb goes out.
Adding another resistor in parallel to an existing circuit:
Decreases the total resistance and increases the total current
Increases the total resistance and decreases the current
Has no effect on the total current
Increases both resistance and current
Each new parallel branch gives current another path, so the equivalent resistance drops and the battery supplies more total current. This is why overloading a household circuit trips the breaker.
How should an ammeter and a voltmeter be connected to measure a resistor?
Ammeter in parallel; voltmeter in series
Both in series
Both in parallel
Ammeter in series (low resistance); voltmeter in parallel (high resistance)
An ammeter must carry the same current as the resistor, so it goes in series and should barely add resistance. A voltmeter compares two points, so it goes in parallel and should draw almost no current.
A current of 2.0 A flows for 10 s. How much charge passes a point in the wire?
5.0 C
0.20 C
20 C
12 C
, so .
A 10 µF capacitor is charged to 5.0 V. How much charge does it store?
2.0 µC
0.5 µC
50 µC
125 µC
. 125 is in microjoules: the stored energy, not the charge.
A wire is stretched to twice its original length. Its volume stays the same. What happens to its resistance?
It becomes 4 times larger
It doubles
It stays the same
It halves
. Doubling L at constant volume halves the cross-sectional area A, so R is multiplied by . Doubling is the tempting answer if you forget that the area shrinks.
A 4 Ω resistor is in series with a parallel pair of 6 Ω and 12 Ω resistors. A 24 V battery drives the circuit. What is the total current?
1.1 A
6 A
2 A
3 A
Parallel pair: . Total = 4 + 4 = 8 Ω, so . 1.1 A adds all three resistors as if they were in series (22 Ω).
In the same circuit (4 Ω in series with 6 Ω ∥ 12 Ω, 24 V battery), how much current flows through the 6 Ω resistor?
3 A
1 A
2 A
4 A
Total current is 3 A, and the parallel pair (4 Ω equivalent) gets 3 × 4 = 12 V. The 6 Ω branch carries 12 / 6 = 2 A and the 12 Ω branch 1 A, which adds back to 3 A. More current takes the lower-resistance path.
