There is a DC power supply whose internal resistance can be neglected to deliver current to the series resistor A. B. when the 90 ohm resistor a is short circuited, the current in the circuit is four times that before the short circuited, then the resistance value of the resistor B is?

There is a DC power supply whose internal resistance can be neglected to deliver current to the series resistor A. B. when the 90 ohm resistor a is short circuited, the current in the circuit is four times that before the short circuited, then the resistance value of the resistor B is?

I = u / r the current is inversely proportional to the resistance
The current is four times the original, and the resistance is one fourth of the original
Rb / (Ra + Rb) = 1/4
The solution is RB = 30 Ω

As shown in the figure, measure the resistance (1) Ra and Rb of the same semicircular metal sheet from different directions to see if they are equal（
Mobile phones can't transmit pictures.

Problems to be considered
(1) Mainly depends on the length of the resistance
(2) Resistance connection, parallel connection or series connection, or partial short circuit
So it's easy to analyze,

As shown in the figure, after two constant resistors R1 and R2 are connected in series, they are connected to a DC power supply whose voltage U is stable at 12V. Someone connects a voltmeter whose internal resistance is not much greater than R1 and R2 to both ends of R1, and the indication of the voltmeter is 8V. If he connects the voltmeter to both ends of R2, the indication of the voltmeter will be ()
A. Less than 4vb. Equal to 4vc. Greater than 4V. Less than 8vd. Equal to or greater than 8V

If the voltmeter is connected to both ends of R1 and the indication of the voltmeter is 8V, then the voltage at both ends of R2 is 4V. When the voltmeter is connected to both ends of R2, because the internal resistance of the voltmeter is not much greater than R1 & nbsp; R2, the resistance of the parallel connection of the voltmeter and R2 is less than R2, and the total voltage of the parallel connection of R1 and R2 is u = 12V, then the voltage of R1 is greater than 8V and the voltage of the circuit of the parallel connection of the voltmeter and R2 is less than 4V

In the circuit shown in the figure, it is known that the internal resistance Ra of the ammeter is many times smaller than the constant resistance R (RA ＜ R). When the key K is open, the indication of the ammeter is 3.0A; when the key K is closed, the indication of the current is 1.0A. If the standard resistance R0 = 6 Ω, the internal resistance Ra of the ammeter is ()
A. 2ΩB. 3ΩC. 12ΩD. 18Ω

It is known from the title that RA < R, the current in the circuit is mainly controlled by R, when the key K is closed, the total current in the circuit is basically unchanged, then when the key K is closed, the current representation is 1.0A, so the current through R0 is I0 = 3a-1a = 2A, which is twice the current of the ammeter. According to the inverse ratio between the current and the resistance of the parallel circuit, RA = 2r0 = 2 × 6 = 12 (Ω), so select: C