In order to solve the teachers in the office when the drinking water problem, the school has a new batch of desktop warm water dispenser, its nameplate is shown in the table below. As shown in the fig

In order to solve the teachers in the office when the drinking water problem, the school has a new batch of desktop warm water dispenser, its nameplate is shown in the table below. As shown in the figure is a simple drinking fountain (1) I=< td>U

P

=2.5A

(2) Q water=mc?t=189000J,

W=Pt=550W×6×60s=198000J,

then η=

QI water

W

×100%=95%

(3) Heating when S is closed, we have:P heating=

;

S is insulated when disconnected, then we have:P insulation=

U2

R2

R2

U2

R1+R2

,

In accordance with the question, there are:P insulation=

1

2

P Heating, substituting values,

Solve:

=

R1

R2

< td>1

1

.

Answer: (1) The current during normal operation is 2.5A; (2) The efficiency of the water dispenser is 95%; (3) The ratio of the two resistors is 1:1.