4V battery with 100 Ω resistance in parallel, the voltage is only 3.9v, what's the reason? Is it related to the current?

4V battery with 100 Ω resistance in parallel, the voltage is only 3.9v, what's the reason? Is it related to the current?

Is there only a 100 ohm resistor connected to the outside of the battery?
If so, the internal resistance of the battery is 4 V / (3.9 V / 100 Ω) - 100 Ω = 2.56 Ω

In the series circuit, R and the sliding rheostat R0 are connected in series. When the voltage at both ends of R0 is 4V and the passing current is 0.2A, calculate the R0 resistance. Change the resistance, and the passing current in the circuit is 0.5A. At this time, R0 consumes 0.5W of electric power, and calculate the rated resistance R resistance

R0 = 4V / 0.2A = 20 Ω
Let the resistance of the sliding rheostat be r '
Then R '= P / I ^ 2 = 2 Ω
At this time, the voltage is 2 * 0.5 = 1V
Yes, R
At the beginning, U-4 / r = 0.2
Later, U-1 / r = 0.5
The solution is u = 6V, r = 10 Ω

As shown in the figure, the voltage at both ends of the power supply does not change, and the resistance value of resistance R1 is 2 Ω. Close the switch S. when the slide P of sliding rheostat is at point a, the indication value of voltmeter V1 is 4V, and that of voltmeter V2 is 10V. When the slide P of sliding rheostat is at point B, the indication value of voltmeter V1 is 8V, and that of voltmeter V2 is 11V. Then the resistance value of resistance R2 is 4V___ Ω．

According to the constant total voltage, the sum of the indication of voltmeter V2 and the voltage at both ends of R1 should be the power supply voltage. The sum of the indication of voltmeter V1 and the voltage at both ends of resistance R2 is also the power supply voltage. According to the equal power supply voltage, the following equation can be obtained: 10V + i1r1 = 11V + i2r1 ①4V+I1R2=8V+I2R2… ② It can be reduced to R1 (i1-i2) = 1V ③R2（I1-I2）=4V… ④ Substituting R1 = 2 Ω into ③ and ④, the solution is R2 = 8 Ω, so the answer is 8 Ω

After a small bulb with rated current of 0.2A and a sliding resistor are connected in series, they are connected to both ends of the power supply with constant voltage of 9V,
A small bulb with rated current of 0.2A and a sliding resistor are connected in series to both ends of the power supply with constant voltage of 9V. After the switch is closed, move the slide. When the power consumed by the sliding resistor is 1.04w, the small bulb just lights up
Q: 1. The rated voltage and power of the small bulb
2. The normal light of the small bulb is the resistance value of the rheostat access circuit
3. Move the sliding rheostat. When the current in the circuit is 0.1A, the actual power of the small bulb is? (regardless of the influence of temperature)

1, rated power of small bulb: 0.2x9-1.04 = 0.76w
Rated voltage: 0.76/0.2 = 3.8V
2.1.04 / 0.2 ^ 2 = 26 Ω
3. Small bulb resistance: 3.8 / 0.2 = 19 Ω
So the power: 19x0.1x0.1 = 0.19w