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Xiao Ming's home is equipped with a TV, a washing machine and two lights. The current under normal operation is 200mA, 1a and 250mA respectively. If the current in the circuit is not allowed to exceed 3a, how many lights of 300mA can he install at most?
Xiaoming's family has a TV, a washing machine and two lights. It is known that the current passing through them is 200mA, 1a, 300mA and 250mA respectively. If the circuit current of Xiaoming's family is not more than 3a, can these appliances be used at the same time?
Because the consumers are all in parallel, their total current is: I = 0.2A + 1A + 0.3A + 0.25A = 1.75A; & nbsp; & nbsp; because 1.75A < 3a, these consumers can be used at the same time. A: these consumers can be used at the same time
Xiao Ming has a refrigerator, a washing machine, an electric fan, and four lights at home‘‘‘‘‘‘‘‘
Xiaoming has a refrigerator, a washing machine, an electric fan, and four lamps at home. The current is 1.1a, 0.89a, 0.59a, and 0.48A, respectively. It is required that the current on the trunk road should not exceed 5A. Now we need to install an air conditioner with the current of 2.2A. What can you do to install it in the circuit?
This is our final examination of the second grade physics, some people say it is to change the fuse and so on‘‘
The answer I did was to connect a line from the power supply in series with the air conditioner. If it's wrong or right, choose one for me and tell me why
Can be installed in the circuit. Reason: the chance of all the above electrical appliances being used at the same time is too small. Even if all the electrical appliances are used at the same time, the maximum current is only 5.26a, which is only a little more than the allowable current in the trunk road
If one-half of a is equal to three-quarters of B, then the ratio of a to B equals () to ()
1 / 2 a = 3 / 4 B
So a = 3 / 2 b
A: B = 3:2
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