Advanced Electricity Variables and Units

A 6 Ω light bulb uses 60 w. How much current does the light bulb use?

6 Ω refers to?
1. q
2. R
3. V
4. I
5. Fe
6. Ef
7. C
8. PEelec
9. P
A 6 Ω light bulb uses 60 w. How much current does the light bulb use?

60 w refers to?
1. q
2. R
3. V
4. I
5. Fe
6. Ef
7. C
8. PEelec
9. P
A 6 Ω light bulb uses 60 w. How much current does the light bulb use?
R = 6 Ω
P = 60 w
You are looking for current. What variable is that?
1. q
2. R
3. V
4. I
5. Fe
6. Ef
7. C
8. PEelec
9. P
A 6 Ω light bulb uses 60 w. How much current does the light bulb use?
R = 6 Ω
P = 60 w
I = ?

Which equation will you use?
1. V = IR
2. I = Q/t
3. 1/Rt = 1/R₁ + 1/R₂ + 1/R₃
4. C = q/V
5. P = I²R
6. P = VI
7. V = PEelec/q
How much charge is necessary to put 12 v across a 3 μF capacitor?

What is 12 v?
1. q
2. R
3. V
4. I
5. Fe
6. Ef
7. C
8. PEelec
9. P
How much charge is necessary to put 12 v across a 3 μF capacitor?
V = 12 v

What is 3 μF?
1. q
2. R
3. V
4. I
5. Fe
6. Ef
7. C
8. PEelec
9. P
How much charge is necessary to put 12 v across a 3 μF capacitor?
V = 12 v
C = 3 μF
What are you looking for?
1. q
2. R
3. V
4. I
5. Fe
6. Ef
7. C
8. PEelec
9. P
How much charge is necessary to put 12 v across a 3 μF capacitor?
V = 12 v
C = 3 μF
q = ?

Which equation will you use?
1. V = IR
2. I = Q/t
3. 1/Rt = 1/R₁ + 1/R₂ + 1/R₃
4. C = q/V
5. P = I²R
6. P = VI
7. V = PEelec/q
How much voltage is necessary to give 36 J to 6 coulombs?

36 J is what?
1. q
2. R
3. V
4. I
5. Fe
6. Ef
7. C
8. PEelec
9. P
How much voltage is necessary to give 36 J to 6 coulombs?
PEelec = 36 J

6 coulombs is what?
1. q
2. R
3. V
4. I
5. Fe
6. Ef
7. C
8. PEelec
9. P
How much voltage is necessary to give 36 J to 6 coulombs?
PEelec = 36 J
q = 6 c

What are you looking for?
1. q
2. R
3. V
4. I
5. Fe
6. Ef
7. C
8. PEelec
9. P
How much voltage is necessary to give 36 J to 6 coulombs?
PEelec = 36 J
q = 6 c
V = ?

Which equation will you use?
1. V = IR
2. I = Q/t
3. 1/Rt = 1/R₁ + 1/R₂ + 1/R₃
4. C = q/V
5. P = I²R
6. P = VI
7. V = PEelec/q