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The resistance of an electrical appliance is 100 ohm, and its maximum allowable current is 100 mA. If it is connected to a circuit with a current of 1.1 A, what method should be adopted?
First of all, we need to understand that the total current is 1.1a, and the current quality energy through a 100 ohm resistor is 100 Ma, which is 0.1 A. what about the current of 1 a? This can only be used in parallel circuit, which has the function of shunt. The current flowing through another current is 1A. When we know the voltage of parallel circuit, u = U1 = U2, that is
When 10 ohm resistance is connected between two points AB, the current indication is 0.5A. If a and B are connected with 20 ohm resistance, what is the current indication?
The circuit diagram is: ab two-point resistance, a constant resistance R1, ammeter series
The answer is greater than 0.25A and less than 0.5A
I don't know why, please answer in detail, thank you!
Very simple, ignoring the internal resistance, calculate the power supply for 5V. That means that this is the minimum value, the actual internal resistance points voltage, the actual power supply voltage is higher than 5V. After changing the resistance, the voltage of 5V is 0.25A, which is not difficult, and the internal resistance of the power supply does not change, so the total current will be larger!
If the electric appliance in the circuit is 60 ohm, 1 / 5 of the total current in the circuit should pass through the electric appliance___ Joint__ Ohmic resistance
If parallel, then I = I1 + I2
So I always = 5i1
Because u = U1 = U2
So r total = 1 / 5R1
And because 1 / R total = 1 / R1 + 1 / R2
R2 = 15 ohm
That is the first answer: and, 15
If in series, then I = I1 = I2
It doesn't match the meaning of the question, so we can only take the first answer
A voltmeter with a range of 0 to 150V has an internal resistance of 20K ohm. Connect it in series with a large resistance at both ends of the 110V circuit. The reading of the voltmeter is 5V. Calculate the size of the external resistance
Don't think about the range of the meter. It's confusing,
The current in the circuit is 5V / 20K Ω
External resistance R = 110 / i-20k Ω = 420k Ω
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