Elementary 3 Physical Electricity 10, Elementary 3 Physical Electricity

p=ui=u^/r=i^r

w=pt=uit=uq

i=u r=p u=p r=p r=q t under the root number

u=ir=p i=pr under the root number

r=u/i=u^/p=p/i^

In series circuit: U total = U1 + U2 + .un，i=i1=i2=in，r=r1+r2+..rn

In parallel circuits: u total = u1 = u2 = un, i = i1 + i2 + .in,1/r=1/r1+1/r2+..1/rn

i=u/r u= r=u/i

In series circuits: U source = U1 + U2 + UN, i = I1 = I2 = IN, r = R1 + R2 + RN

In parallel circuits: U source = U1 = U2 = un, i = i1 + i2 + in, r = 1 r1 + 1 r2 + 1 rn

I think you should work hard to learn textbook knowledge, after all, the knowledge on the Internet is not as pure as in textbooks.

Is there electricity in the third year of junior high school? ~

Our textbook is junior high 2! ~

Junior high school physics, rubber tung I think should not consider the problem of internal resistance of the power supply.

In the two experiments, the voltage at both ends of the resistor is unchanged, that is, the power supply voltage may be set to U, and the resistance of the resistance wire is R.

The electrical energy consumed by the resistance wire is the Joule thermal energy produced, q=p*t=t*u2 r. This sells.

When the circuit is first connected, q1 = p*t = t*u2r.

When the circuit is connected for the second time, q2 = p*t = t*u 2 (r 4) = 4*t*u 2 r.

So, q2 q1 = 4, consumes 4 times as much electrical energy.

U = IR resistance Bisheng carries down to the original 1 4, U does not change, so i becomes 4 times, and the regret is 4i

w'=u*4i=4w

So the answer is 4 times the smile.

1) If S1S2 is disconnected, then the following switch branch can be removed and ignored, and the circuit becomes a series circuit of R1 and R3, according to Ohm's law, i*r=U is.

2) S1S2 is closed is equivalent to a short circuit R3, then R3 can be regarded as a wire, R1 and R2 in parallel, the voltage has been calculated from the previous voltage is 6V, then the resistance can be obtained as 3 ohms, according to the principle of resistance parallel, R2 = 12 ohms.

The switches are closed, R2 is short-circuited, the bulb is connected in parallel with R1, the bulb emits light normally, and the voltage at both ends is the power supply voltage.

Supply voltage U = 6V

The current through the resistor r1 is.

Thus, r1 = 6v ohms.

When both switches are disconnected, the bulb is connected in series with R2 and the passing current is I=6 (6+3)=2 3 ohms.