The full scale current of a Ma meter is 1mA, the internal resistance of the meter is 98 Ω, and the shunt resistance is 2 Ω

The full scale current of a Ma meter is 1mA, the internal resistance of the meter is 98 Ω, and the shunt resistance is 2 Ω

1. When the meter head is full scale, the terminal voltage is u = I1 * r = 1 * 98 = 98mv
2. When the meter head of the milliammeter is in full scale, the current I2 on shunt resistance R = u / r = 98 / 2 = 49ma
3. When the meter head is full scale, the upper limit of measurement is I = I1 + I2 = 1 + 49 = 50mA

The ammeter is actually a meter, (the maximum allowable current of the meter is 0.6A, and the internal resistance is ignored)
And resistance (R1 = 5 Ohm) in series, it will be improved to range 3A ammeter, need to parallel a resistance? Do have to explain, thank you

The voltage at both ends of the parallel circuit is equal
When it is improved to an ammeter with a measuring range of 3A
The maximum current of the meter is I1 = 0.6A
The maximum current of parallel resistance is I2 = 2.4a
I1r1 = i2r2, R2 = 1.25 Ω

When measuring AC voltage, the current flowing through the meter head is ()
(A) AC current (b) constant DC current (c) DC current (d) pulse current

(C) Direct current

When measuring the voltage, why should the voltmeter be connected in parallel with the circuit under test

Because the internal resistance of the voltmeter is very large, if you connect the voltmeter in series to the circuit under test, it is almost equivalent to the circuit under test open circuit, and the measured voltage is almost the power supply voltage; if the voltmeter is connected in parallel with the circuit under test, because the internal resistance of the voltmeter is very large, the actual resistance after parallel connection is basically equivalent to the resistance of the part under test, and the measured voltage is basically the voltage at both ends of the circuit under test