Heating a pot of water with an electric heater with resistance R1 takes t1min to boil, and heating the same pot of water with an electric heater with resistance R2 takes t2min to boil. The time ratio of heating the same pot of water with two resistors in series to heating the same pot of water with two resistors in parallel is () and the power supply voltage remains unchanged A t1+t2/t1t2 B (T1 + T2) square / t1t2 C t1t2/t1+t2 D T1 + T2 / (T1 + T2) square

Heating a pot of water with an electric heater with resistance R1 takes t1min to boil, and heating the same pot of water with an electric heater with resistance R2 takes t2min to boil. The time ratio of heating the same pot of water with two resistors in series to heating the same pot of water with two resistors in parallel is () and the power supply voltage remains unchanged A t1+t2/t1t2 B (T1 + T2) square / t1t2 C t1t2/t1+t2 D T1 + T2 / (T1 + T2) square


Suppose the mass of water is m, the specific heat capacity is C, and the power supply voltage is u, then u ^ 2 / R1 = CMT1, u ^ 2 / r2 = CMT2, so R1 = u ^ 2 / (CMT1), R2 = u ^ 2 / (CMT2), and the total resistance after series connection is r = R1 + R2 = u ^ 2 / (CM) * [(T1 + T2) / (t1t2)], so the total resistance after parallel connection is r = R1 * R2 / (R1



The specification of an electric heater is 220V, 880W. What is the resistance of the electric heater when it works normally


R=P/U*U



I have 300 ohm, 200 ohm, 510 ohm and 470 ohm resistance, but I need 1000 ohm resistance
How to connect these four kinds of resistors to get 1000 ohm resistance


300 series 300 Series 200 Series 200 = 1000
200 in parallel 200 in series 300 in series 300 in series 300 = 1000
wait