11th grade Current, voltage and resistance worksheet

Current, voltage and resistance

Grade 11 Science Worksheet - Challenge Me

Circle the best answer, or write your answer on the line.

1. 1.

   Ari's small bulb carries 0.2 A at 2 V and 0.5 A at 7 V. By what percent does its resistance rise?
2. 2.

   Bo's small bulb carries 0.25 A at 1.5 V and 0.4 A at 6 V. By what percent does its resistance rise?

3. Read this first

   Readings for a small bulb (example data)

   voltage (V)	current (A)
   1	0.1
   3	0.2
   6	0.3
   9	0.36

4. 3.

   Jun reads the bulb table. How does the bulb's power change when the voltage goes from 3 V to 9 V?

   • It rises 9 times, since P = V² ÷ R and the voltage triples
   • It rises 5.4 times, from 0.6 W to 3.24 W
   • It rises 3 times, in step with the voltage
   • It rises 1.8 times, in step with the current
5. 4.

   Kira sees the bulb's current-voltage curve flatten as voltage rises. What does it show?

   • The curve is not a line, so V = IR cannot be used for the bulb
   • The slope falls, so the resistance falls
   • The slope ΔI ÷ ΔV falls from 0.05 to 0.02 A per V, so resistance rises
   • The slope stays near 0.1 A per V, so the resistance is steady
6. 5.

   Bulb A (6 Ω) is in series with bulbs B and C (12 Ω each, in parallel) across 12 V. Which glows brightest, and with what power, Luz?

   • The three glow equally, since one battery powers them
   • A: 24 W against 12 W for B and C, from V² ÷ R with 12 V
   • A: 6 W, against 3 W for each of B and C
   • B and C: the larger resistances, 12 Ω each, glow brightest
7. 6.

   “Circuit schematic (example data):” Milo says the bulb uses up current, so less leaves the bulb than enters it. Which reply fits the schematic?

   • No: the ammeters agree because they sit before the switch
   • No: the bulb adds current as it heats up
   • Yes: the bulb shines, so some of the current must be used up inside it
   • No: both ammeters read 0.30 A; the bulb transfers energy, not current
8. 7.

   Pia says a 1.5 V battery is safe across a thick wire of 0.05 Ω because 1.5 V is a small voltage. Which reply fits Ohm's law?

   • Yes: 1.5 V ÷ 0.05 Ω is a tiny current, 0.03 A
   • No: 1.5 V ÷ 0.05 Ω = 30 A, which would heat the wire
   • Yes: a battery this small has too little voltage to heat a wire
   • No: 1.5 V × 0.05 Ω = 0.075 A, which is already too much
9. 8.

   A 12.0 V battery with 0.50 Ω internal resistance is joined to a 5.5 Ω heater. What voltage is across the heater, Ren?

   • 10.9 V: 12 − (12 ÷ 5.5) × 0.50
   • 1.0 V, the voltage lost inside the battery
   • 11.0 V: 12 ÷ 6.0 = 2.0 A, then 2.0 × 5.5
   • 12.0 V, the emf of the battery
10. 9.

    The 12.0 V battery with 0.50 Ω internal resistance drives a 5.5 Ω heater at 2.0 A. How much power is wasted inside the battery, and how much reaches the heater, Tala?

    • 2.0 W is wasted inside; the heater gets 24 W
    • 2.0 W is wasted inside; the heater gets 22 W, about 92% of 24 W
    • 1.0 W is wasted inside; the heater gets 23 W
    • 2.0 W is wasted inside; the heater gets 22 W, about 8% of 24 W
11. 10.

    An identical bulb B is added to bulb A on a 6 V battery. How does A's brightness change when B is added in parallel, and when B is added in series, Zee?

    • Parallel: A falls to half. Series: A falls to half
    • Parallel: A falls to half. Series: A falls to a quarter
    • Parallel: unchanged. Series: A's power falls to a quarter
    • Parallel: unchanged. Series: A falls to half