If the two resistors R1 (10 Ω, 0.5A) and R2 (15 Ω, 0.4A) are connected in parallel, the maximum current in the trunk circuit is () a, There are two existing resistors R1 (10 Ω, 0.5A) and R2 (15 Ω, 0.4A). If the two resistors are connected in parallel, the maximum current in the main circuit is () A. if the two resistors are connected in series, the maximum power supply voltage is () v

If the two resistors R1 (10 Ω, 0.5A) and R2 (15 Ω, 0.4A) are connected in parallel, the maximum current in the trunk circuit is () a, There are two existing resistors R1 (10 Ω, 0.5A) and R2 (15 Ω, 0.4A). If the two resistors are connected in parallel, the maximum current in the main circuit is () A. if the two resistors are connected in series, the maximum power supply voltage is () v

Parallel connection means that the U of the two resistors are the same, so u1i2, in order to prevent R2 from burning out, I1 = I2 = 0.4A. Then u = I (R1 + R2) = 0.4 * 25 = 10V

If two resistors are connected in parallel and R1: R2 = 1:2, the current ratio I1: I2 =, the voltage ratio U1: U2 =, and the electric power ratio P1: P2 =

I1:I2=(2:1),U1:U2=(1:1),P1:P2=(2:1)
Parallel equal voltage, U1: U2 = 1:1
And U1 = I1 * R1, U2 = I2 * R2
So I1 * R1 = I2 * R2
I1/I2=R2/R1=2:1
P=UI
P1/P2=(U1*I1)/(U2*I2)=I1/I2=2:1

Two resistors R1 and R2 in series in the circuit, known as R1 = 10 Ω, R2 = 30 Ω, U1 = 10V, calculate R2 to get current I2 and U2

10V / (10 + 30) = 0.25A (total current of the circuit). The voltage of resistance R2: 0.25A * 30 = 7.5V

Two resistors R1: R2 = 2:1, connected in series in the circuit, I1: I2 = (), U1: U2 = (); connected in parallel in the circuit, I1: I2 = (), U1: U2 = ()

According to Ohm's Law: I = u / R, u = I * r. in series circuit, because I = I1 = I2, so U1 / ri = U2 / R2, then U1 / U2 = R1 / r2 = 2:1; in parallel circuit, because u = U1 = U2, so I1 * R1 = I2 * R2, then I1 / I2 = R2 / R1 = 1:2