Two resistors R1 = 4 Ω, R2 = 8 Ω. If they are connected in series in the circuit, the current passing through them is L1 and L2 respectively If the voltages at both ends are U1 and U2, the ratios of L1 to L2 and U1 to U2 are

Two resistors R1 = 4 Ω, R2 = 8 Ω. If they are connected in series in the circuit, the current passing through them is L1 and L2 respectively If the voltages at both ends are U1 and U2, the ratios of L1 to L2 and U1 to U2 are

The two resistors are connected in series, and the current is equal, that is, I1: I2 = 1:1
U1:U2=R1:R2=1:2

Find the resistance R1 / 220 Ω R2 / 2.2 Ω R3 / 1 Ω R * / 39 Ω of tda2003 power amplifier circuit diagram. What is their actual resistance?
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1. We can't see your schematic,
The resistance value of the peripheral resistance of 22030 is not fixed, and the resistance value of its negative feedback circuit determines its magnification,
3. You can change the resistance between 2 and 4 pins by referring to other 2030 circuits. The 2.2 ohm one you mentioned can not move. This is to eliminate self excitation,

In the circuit diagram, R1 = 10 ohm, R2 = 20 ohm are connected in parallel. After the switch is closed, the current expression is 0.3 ampere. The resistance is R1 voltage? R2 current?

Because the two resistors are in parallel, the voltage of R1 and R2 are the same. The resistance of trunk road r = (10 × 20) / (10 + 20) = 20 / 3 Ω, the voltage U = IR = 0.3 × 20 / 3 = 2V, that is, the voltage of R1 is 2V, and the current of R2 I = u / r2 = 0.1A