Resistor R1 is connected in series in the circuit. It is known that the voltage at both ends of R1 is 1.5V, the voltage at both ends of R2 is 405v, and R1 + R2 = 600

Resistor R1 is connected in series in the circuit. It is known that the voltage at both ends of R1 is 1.5V, the voltage at both ends of R2 is 405v, and R1 + R2 = 600

In case of series connection, the resistance is proportional to the voltage, then:
R1/R2=1.5/405
And R1 + R2 = 600
R1, R2 are obtained

In the circuit as shown in the figure, the resistance value of resistance R1 is 5 Ω, the current representation number is 0.6A, and the voltage at both ends of resistance R2 is 2.5V. Calculate: (1) the current passing through R1 (2) the current passing through R2 (3) the resistance of R2 (4) the total resistance in the circuit

It can be seen from the circuit diagram that R1 and R2 are connected in parallel, and the ammeter measures the main circuit current. (1) ∵ the voltage at both ends of each branch in the parallel circuit is equal, ∵ U1 = U2 = 2.5V, the current passing through R1: I1 = u1r1 = 2.5V, 5 Ω = 0.5A; (2) ∵ the main circuit current in the parallel circuit is equal to the sum of the branch currents, ∵ the current passing through R2: I2 = i-i1 = 0.6a-0.5a = 0.1A; (3) the resistance of R2: R2 = u2i2 = 2.5V, 0.1A = 25 Ω; (4) the resistance of R2: R2 = u2i2 = 2.5V, 0.1A = The total resistance in the circuit: r = UI = 2.5v0.6a ≈ 4.2 Ω. Answer: (1) the current through R1 is 0.5A; (2) the current through R2 is 0.1A; (3) the resistance of R2 is 25 Ω; (4) the total resistance in the circuit is about 4.2 Ω

In the circuit as shown in the figure, it is known that the power supply voltage is 4.5V, the current of R1 is 0.2A and the voltage is 2V; the current, voltage and resistance of R2 are calculated

It can be seen from the circuit diagram that two resistors are connected in series, ∵ the current in the series circuit is equal, ∵ the current through R2: I2 = I1 = 0.2A, ∵ the total voltage in the series circuit is equal to the sum of the partial voltages, ∵ the voltage at both ends of R2: U2 = u-u1 = 4.5v-2v = 2.5V, obtained from I = ur, the resistance of R2: R2 = u2i2 = 2.5v0.2a = 12.5

When the voltmeter is connected with a resistor R1 in series, the maximum allowable voltage at both ends of the circuit is 5V, and the maximum allowable voltage at both ends of the circuit is 6V when the voltmeter is connected with another resistor R2 in parallel. At this time, the maximum voltage at both ends of the circuit is 5V
The range of voltmeter is 0-3v!

The range of voltmeter is 3V. When R1 is connected in series, R1 voltage is 2V. R1 = RV * 2 / 3
When R2 is connected in series, R2 voltage is 3V, R2 = RV, the maximum current allowed by Voltmeter is the same twice, so: after R1 and R2 are connected in parallel, RV * 3 / 2 and RV are connected in parallel, and the resistance after connection in parallel is RV * 2 / 5, that is, the parallel resistance is 2 / 5 times of RV, so the maximum voltage is 3V + 3V * 2 / 5 = 4.2V