The resistance ratio of the two resistance wires is 3:2, and the current ratio through them is 1:4. If they generate the same heat, the power on time ratio is 3:2——

The resistance ratio of the two resistance wires is 3:2, and the current ratio through them is 1:4. If they generate the same heat, the power on time ratio is 3:2——

Q1=I1²R1t1
Q2=I2²R2t2
And Q1 = Q2
That is, I1 & # 178; r1t1 = I2 & # 178; r2t2
t1：t2=(I2/I1)²(R2/R1)=(2/3)²(4/1)=16:9
A: slightly

There is an ammeter g with internal resistance of RG = 100 and partial full current of Ig = 3mA. It is refitted into a voltmeter by connecting a 900 resistor in series. What is the range of the voltmeter

R=Rg+R0=100Ω+900Ω=1000Ω
U=I*R0=0.003A*1000Ω=3V

1. A voltmeter is composed of galvanometer g and resistance R in series. Why is the reading of the voltmeter smaller than the accurate value, which means that the current flowing through the voltmeter is smaller and the resistance in series is larger?
2. If it is found that the reading of a modified ammeter is always slightly smaller than the accurate value, why should a resistor be connected in series? Why should the resistor be very small?

Because the reading of the voltmeter is determined by the deflection angle of the meter head, and the deflection angle of the meter head is proportional to the current flowing through the meter head, the reading of the voltmeter is smaller than the accurate value, which means that the current flowing through the meter head is smaller and the voltage of the meter head is smaller, which means that the partial voltage effect of the resistance R is larger, so the resistance R is larger

A voltmeter is composed of ammeter g and resistor R in series. If you find that the reading of this voltmeter is a little smaller than the accurate value in use
A voltmeter is composed of ammeter g and resistor R in series. If it is found that the reading of the voltmeter is always a little smaller than the accurate value in use, which measures can be taken to improve it?
C has a much smaller resistor in parallel with R
Why not? If the resistance at both ends of R is much smaller than R, the total resistance after parallel connection will be much smaller than R. if a much larger resistance is paralleled on R, the total resistance after parallel connection will be slightly smaller than R?
Help me explain my gratitude!

Let's look at the parallel formula r = R (1) r (2) / {R (1) + R (2)}
Suppose that R (1) is already above. R (2) is the last addition
Then I divide the numerator and denominator by R (2) r (total) = R (1) / {[R (1) / R (2)] + 1}
See the denominator? How to calculate 1 + a number larger than zero is greater than 1
R (1) / R (2) should be smaller. As long as the denominator is larger, give me a score if you think it's right. Thank you! If R (2) is smaller, then R (1) / R (2) is larger than 1. 1 + is larger than 1
It is more than 2. Then R (total) will be expanded by more than 2 times. Then the range of voltmeter will be expanded by more than 2 times