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Connect two small bulbs L1 and L2 marked with 6v3w and 6v9w in series and connect them to 8V power supply to calculate the actual power of L1 and L2
P = u / I 6v3w standard current 0.5A, resistance 126v9w standard current 1.5A, resistance 4 series to, total resistance 12 + 4 = 16, total current I = u / r = 8 / 16 = 0.5a0.5a, through 12 ohm 6v3w generated 6V voltage 0.5A current, through 4 ohm 6v9w generated 2V voltage (2 + 6 just equal to 8 series
If the bulb L1 marked with 12v6w and the bulb L2 marked with 6v3w are connected in parallel, then the L1 and L2 are connected to the 6V power supply. What is the actual power of L1 and L2?
L1 and L2 are connected to 18V power supply in series. What is the actual power of L1 and L2?
Because two lamps are connected in parallel, U1 = U2 = 6V
Because U2 real = U2 amount = 6V
So P2 real = P2 amount = 3W
And P1 = P2 / 4 = 1.5W
For reference only!
In the circuit shown in the figure, L1 and L2 are marked with 6v3w and 6v2w respectively. When switch S is connected to a, L1 will light normally, and when switch S is connected to B, L2 will light normally. (1) setting
As shown in the figure, L1 and L2 are marked with 6v3w and 6v2w respectively. When switch S is connected to a, L1 will light normally, and when switch S is connected to B, L2 will light normally
(1) What is the resistance value of the constant resistance R?
(2) What is the power supply voltage?
3. What is the ratio of the two electric powers
1) When switch S is connected to a: UL = 6V
Then I = P / u = 3 / 6 = 0.5A
When switch S is connected to B: UL = 6V
I=P/U=2/6=1/3
Isn't that right?
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