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Do you need to adjust the ohm to zero after measuring the resistance of 4 ohm and 5 ohm with the ohm gear of the multimeter Do you need to adjust the ohm to zero after measuring the resistance of 4 ohm and 5 ohm respectively with the ohm gear of the multimeter
Electronic digital meter doesn't need to adjust zero when measuring resistance, and there is no device to adjust zero on the meter. Pointer meter generally doesn't need to adjust zero again because 4 ohm and 5 ohm are measured in the same gear. However, if the battery is old and measuring small resistance costs more electricity, the zero point may move. You can short circuit the needle to determine whether the zero point has moved, so as to decide whether it needs to be adjusted
Connect a bulb to the power supply, its electric power is 9W, and then take a resistor in series with it on the same power supply, the electric power of the bulb becomes 4W, calculate the electric power of the resistor
The total voltage of two connection methods is equal, IR = I '(R + R')
Then we use p = I square r to express I and I '
R = 2R '
In the series circuit, the actual power is proportional to the resistance, so the power of the resistance is 2W
Connect a 6V 3W bulb in series with an 18 ohm resistor to a 12V power supply. What is the current in the circuit
The specification of a color TV set is 50V 25W. To be connected in a 220V circuit, at least several such bulbs need to be connected in series
The resistance of 6V 3W bulb is r = 6 ^ 2 / 3 = 12, so the total resistance of the circuit is 30 and the current is 12 / 30 = 0.4A
A bulb marked "12V 6W" and a 20 ohm resistor are connected in series to the power supply with voltage of U,
1. A bulb marked with "12V 6W" and a 20 ohm resistor are connected in series on the power supply with voltage of u, and the bulb can light normally. From this, we can know the power supply voltage of u and the electric power consumed by the power supply. 2. The voltage at both ends of a circuit is 20V, and the electric energy consumed within 4S is 40J, so what is the electric power? What is the current passing through? What is the resistance of this current?
The current through the bulb is I = P / u = 6 / 12 = 0.5A
The voltage at both ends of the resistor ur = IR = 0.5 × 20 = 10V
Power supply voltage U = u + ur = 22V
Power consumption P = UI = 22 × 0.5 = 11W
From w = Pt, P = w / T = 40 / 4 = 10W
Current I = P / u = 10 / 20 = 0.5A
The resistance of this circuit is r = u / I = 20 / 0.5 = 40 Ω
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