Question 83: A "220 V, 1000 W" electric furnace, without adding other materials and equipment, it is converted into a 110 V circuit and used, with a power of 1000 W. Among the following measures, the feasible one is ( )
A. Cut the electric furnace wire into two equal lengths, connect it in parallel and then connect it to a 110 V circuit
B. Cut the electric furnace wire halfway and connect it to a 110 V circuit
C Cut the electric furnace wire 3/4, and connect the remaining 1/4 to a 110 V circuit
D. Cut the electric furnace wire 1/4, and connect the remaining 3/4 to a 110 V circuit
D. Cut the electric furnace wire 1/4, and connect the remaining 3/4 to a 110 V circuit 110 V circuit 1/4, and connect the remaining 3/4 to a 110 V circuit 110 V. Circuit
parsing:
1 From the question, we can see that
1.1 The rated voltage of the electric furnace U rating = 220 V,
1.2 The actual circuit voltage U tang = 110 V,
Therefore, we can see that U amount = 2U real:
1.3 requires that the power of the electric furnace remain unchanged after modification,
that is P amount = P real = P real = P 1
2 is obtained from the definition of electric power P = U²/R,
R = U²/P,
R = U²/P,
R = U²/P real,
R = U²/P real,
R = U²/P real,
So R 1/R real = (U²/U real)²=4/1
So R 1 = 1/4 R 1.
3 Note that R here is actually the resistance of the modified electric furnace wire. The current problem is transformed into: as long as the resistance of the modified electric furnace wire is equal to 1/4 of the resistance of the original electric furnace wire, that is, R is actually = 1/4 R.
From the figure above, we can see that for item A, R is real = 1/4 R amount,
for item B, R is real = 1/2 R amount,
for item C, R is real = 1/4 R amount,
for item D, R is real = 3/4 R amount,
, so both items A and C are correct.
summary: "Control variable method" is a commonly used research method in physical experiments. This method can also be used in this question to analyze and discuss in the work
