Ohms Law Worksheet - Fill and Sign Printable Template Online - Free Printable

Let’s go through each question step by step. I’ll explain how to solve them using simple ideas and formulas you’ve learned about power, Ohm’s Law, and energy.

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1. State Ohm’s Law:

Ohm’s Law says:
Voltage = Current × Resistance
Or in symbols: V = I × R

This means if you know two of these values, you can find the third.

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2. Give the units for: V ___ ; I ___ ; R ___

- V (voltage) → volts (V)
- I (current) → amperes or amps (A)
- R (resistance) → ohms (Ω)

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3. Restate these units, using J, s and/or C:

We need to think about what voltage, current, and resistance mean in terms of energy (joules), time (seconds), and charge (coulombs).

- Voltage (V) = energy per charge → J/C
- Current (I) = charge per time → C/s
- Resistance (R) = voltage / current → (J/C) ÷ (C/s) = J·s / C²

So:
- V = J/C
- I = C/s
- R = J·s / C²

*(Note: This is more advanced — but it shows how units connect!)*

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4. State the electric power formula:

Power = Voltage × Current
In symbols: P = V × I

You can also write it as P = I²×R or P = V²/R — but the basic one is P = VI.

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5. Give the unit: P ___ . Restate, using J, s and/or C.

Unit of power is watt (W).

Since power = energy / time → W = J/s

Also, since P = V×I, and V = J/C, I = C/s →
P = (J/C) × (C/s) = J/s → same thing!

So:
P = W = J/s

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6. Another formula for calculating power is: P = ___

Another common formula is:
P = I² × R
(or sometimes P = V² / R — both are correct depending on what you know)

But since they say “another formula”, and we already have P = VI, this is likely asking for P = I²R

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7. Rearrange the formula to solve for Energy: E = ___

We know: Power = Energy / Time → P = E / t

So rearrange:
E = P × t

Energy = Power × Time

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8. The kilowatt hour (kWh) is a unit for ___ which Power Companies use...

kWh is a unit for energy.

Why not watt-second? Because watt-second is too small for big electricity bills. A kWh is 1000 watts used for 1 hour — much bigger and easier to measure household usage.

Example: If you run a 1000W heater for 1 hour → that’s 1 kWh.

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9. What is the resistance of an electric frypan that draws 11 amps when connected to a 110 V circuit? What is the power of the frying pan?

Use Ohm’s Law first:
V = I × R → so R = V / I
R = 110 V / 11 A = 10 Ω

Now power: P = V × I = 110 × 11 = 1210 W

✔ So: Resistance = 10 Ω, Power = 1210 W

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10. If a 120 V line to a socket is limited to 15 A by a fuse, will it operate a 1200 W dryer without blowing the fuse?

Find how much current the dryer needs:
P = V × I → I = P / V = 1200 W / 120 V = 10 A

Fuse allows up to 15 A → 10 A < 15 A → YES, it will work without blowing the fuse.

✔ Answer: Yes

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11. If the power company charges us 8 cents/kw-h for electricity, what does it cost to operate the 1200 W hair dryer for 15 minutes?

First, convert everything to right units:

- Power = 1200 W = 1.2 kW
- Time = 15 minutes = 15/60 hours = 0.25 hours

Energy used = Power × Time = 1.2 kW × 0.25 h = 0.3 kWh

Cost = 0.3 kWh × 8 cents/kWh = 2.4 cents

✔ Answer: 2.4 cents

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12. If the power company sells electrical energy at 11 cents/kWh, how much does it cost to run a 100 W radio for 10 hours?

Power = 100 W = 0.1 kW
Time = 10 hours
Energy = 0.1 × 10 = 1 kWh
Cost = 1 × 11 cents = 11 cents

✔ Answer: 11 cents

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13. What is the resistance of a 120W light bulb running on a 120 V circuit?

Use: P = V² / R → so R = V² / P

R = (120)² / 120 = 14400 / 120 = 120 Ω

✔ Answer: 120 Ω

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Final Answer:

1. V = I × R
2. V: volts (V); I: amperes (A); R: ohms (Ω)
3. V = J/C; I = C/s; R = J·s/C²
4. P = V × I
5. P: watt (W); W = J/s
6. P = I² × R
7. E = P × t
8. Energy — because watt-second is too small for billing; kWh is practical for homes
9. Resistance = 10 Ω; Power = 1210 W
10. Yes (dryer uses 10 A, fuse allows 15 A)
11. 2.4 cents
12. 11 cents
13. 120 Ω