As shown in the figure, the power supply voltage remains unchanged at 9 v. the resistance of R0 is 16.5 Ω. There are two resistors in the dotted line box in the figure, with R1 = 6 Ω and R2 = 2.5 Ω (1) When the switches S1 and S2 are closed and S3 is opened, the total power of the circuit is 3.6 W, and the current through R0 can be calculated. (2) when S1 is opened and S2 and S3 is closed, the indication of the ammeter is 0.36 A. please draw the connection of R1 and R2 into the circuit in the dotted line box through calculation. (3) when S1, S2 and S3 are closed at the same time, the electric power of the resistance in the dotted line box can be calculated

As shown in the figure, the power supply voltage remains unchanged at 9 v. the resistance of R0 is 16.5 Ω. There are two resistors in the dotted line box in the figure, with R1 = 6 Ω and R2 = 2.5 Ω (1) When the switches S1 and S2 are closed and S3 is opened, the total power of the circuit is 3.6 W, and the current through R0 can be calculated. (2) when S1 is opened and S2 and S3 is closed, the indication of the ammeter is 0.36 A. please draw the connection of R1 and R2 into the circuit in the dotted line box through calculation. (3) when S1, S2 and S3 are closed at the same time, the electric power of the resistance in the dotted line box can be calculated

(1) When the switches S1 and S2 are closed and S3 is opened, the total power of the circuit is 3.6 watts, and the current passing through R0, I = Pu = 3.69 = 0.4A; a: the current passing through R0 is 0.4A; (2) when S1 is opened and S2 and S3 is closed, the ammeter's indication is 0.36 a, r = UI = 90.36 = 25 Ω, so the electricity in the dotted line box

Look at the circuit diagram, do your own power amplifier, and mark the size of some components on the circuit diagram, such as resistance, capacitance, diode, etc!
In the circuit diagram, we only know the resistance, where to install the components, and how to know the parameters of those components, such as how to calculate,

The parameter you mentioned should refer to the power of the resistor. If so, strictly speaking, it is necessary for you to know the function of each resistor in the circuit, the voltage at its two ends, and then calculate its power. But generally speaking, in low-voltage (below 12V) circuits, you can choose quarter watt or eighth watt, If the resistance value is small, the protection resistance should be larger

As shown in the figure, the resistance of bulb L1 is 6 Ω and that of I2 is 4 Ω. When S1 is closed and S2 is open, the reading of ammeter is I1 = 0.8A; when S1 is open, S2 is closed
The reading of ammeter is I2 = 0.5A, calculate the resistance and power supply voltage of bulb L3.

When S1 is closed and S2 is open, L1 and L2 are connected in series, and the current is I = 0.8A,
Power supply voltage U = IR = I (R1 + R2) = 0.8A * (6 Ω + 4 Ω) = 10V
When S1 is open, S2 is closed, L1 and L3 are connected in series, and the total resistance is r = u / I = 8V / 0.5A = 16 Ω
The resistance of bulb L3 R3 = r-r1 = 16 Ω - 6 Ω = 10 Ω