I always =Er+R0+R and = you 8=0.52 you A,

According to the shunting principle: I VI =I Total R0R0+R VI = you 22a = 0.26A,

(2) The solution is incorrect. 2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p

The resistance of the bulb is not a fixed value, and the actual power after being connected to the circuit should be solved by I-U&; 2b3p graphic line,

It is correct to take the intersection of two graphs.

Equivalent electromotive force E'= 60v, internal resistance r' = R0+r = 60ω,

I-U diagram of equivalent power supply, as shown in Figure 6&; 2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p

The coordinates of the reading intersection point are (2V, 0.5A),

The actual power of the bulb is p = UI = 2.2w. &; 2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p& amp2b3p

(3) let the voltage at both ends of the bulb be U, wherein the current of I bulbs is I,

Ohm's law of closed circuit: E=U+2I(R0+r),

Substitute the data to get the functional equation: I=0. You -0.0 you u,

Draw the corresponding I-U diagram, as shown in Figure 2, and read out the coordinates of the intersection point as (6.9V, 0.26A).

Actual power P=UI=0. You are 8W.