It is known that R1 = 10 Ω, R2 = 20 Ω. If they are connected in series and connected to a 6V power supply, the power consumption of resistor R1 and R2 per minute is If they are connected to the same power supply, the work done by the current to r1r2 per minute is

It is known that R1 = 10 Ω, R2 = 20 Ω. If they are connected in series and connected to a 6V power supply, the power consumption of resistor R1 and R2 per minute is If they are connected to the same power supply, the work done by the current to r1r2 per minute is

Because it's a series circuit
So r = R1 + R2 = 30 Ω
The total current is equal to U / r = 0.2A
Because all external currents in the series circuit are equal, I1 = I2 = 0.2A
From w = I ^ 2rt, we can get W1 = 24J, W2 = 48j
If it's parallel
From w = (u ^ 2 / R) t, we can get W1 = 216j, W2 = 108j

Resistor R1 is 40 Ω, R2 is 9 Ω in series in 18 V circuit 1. The electric power of R1 is 2, R2 consumes electric energy 3. The work done by current to the whole circuit

1,P1=R1*I²=40*（18/49）²=5.398W
2,W2=P2*t=R2*I²*t=9*（18/49）²*（5*60）=364.35J
3,P=U²/（R1+R2）=6.6W

Two resistors with resistance values of R1 = 40 Ω and R2 = 20 Ω are connected in series and connected to the power supply. If the current intensity through R1 is 0.2A, the current intensity through R2 is______ A. If the voltage of the power supply is not changed and R1 and R2 are connected in parallel to the original power supply, the current intensity through R1 is 0______ A. The current intensity through R2 is______ A．

(1) When two resistors are connected in series, the current in each part of the circuit is equal, so the current intensity through R2 is 0.2A, the voltage of power supply u = I (R1 + R2) = 0.2A × (40 Ω + 20 Ω) = 12V; (2) when two resistors are connected in parallel, the voltage at both ends of two resistors is 12V, then I1 = ur1 = 12v40 Ω = 0.3A, I2 = UR2 = 12V20 Ω = 0.6

There is a circuit, resistance R and small lamp L are connected in series at both ends of the power supply. The power supply voltage is 15V, the known resistance R = 3 Ω, and the electric heat generated by R within 1s in the circuit is 27j, and the rated power of lamp L is 32W

(1) According to q = i2rt, the circuit current is: I = QRT = 27j3 Ω× 1s = 3A, according to I = ur, the voltage at both ends of resistance R is: u = IR = 3A × 3 Ω = 9V, the voltage at both ends of bulb is: UL = u-ur = 15v-9v = 6V, according to I = ur, the bulb resistance RL = Uli = 6v3a = 2 Ω; (2) according to P = u2r, the rated voltage of bulb is: UL = PL, RL = 32W × 2 Ω = 8V; (3) the actual power of small bulb in the circuit is: P = i2rl= (3a) 2 × 2 Ω = 18W; answer: (1) the resistance value of the small bulb is 2 Ω. (2) the rated voltage of the small bulb is 8V. (3) the actual power of the small bulb in the circuit is 18W