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Anonymous users2024-02-051.The two 4 ohms below 6 ohms are connected in parallel to make 2 ohms, and then connected with 6 ohms in series to make 8 ohms. 8 ohms and 24 ohms in parallel is 8*24 (8+24)=6 ohms. 6 ohms and 4 ohms in tandem for 10 ohms.
It is known that the power consumption through a 15 ohm resistor is 15 W, which can be seen by U 2 R=P, U = 15V. It can be seen that the passing current of 10 ohms is 15 10=. So the current through r is 1+.
From KVL, it can be seen that the voltage applied to R is 25-15=10V.
25-15) Euros.
2.For R2, R3 and the voltage source train KVL, clockwise and assume that the positive resistors are in the opposite direction of the arrow.
12-v2+v3=0
where v2 = 10 * i1 and v3 = 5 * i2
Considering that 3+i1+i2=0, the above two formulas are obtained by i1=, i2=
To find the output power, you can use p=iu for the power supply.
For the current source, the Kvl of the R1 and R2 columns has V+3*25+10*, and V=93V, so P 93*3=279W
In addition to i2, the voltage source also has a current i3 for r4, i3 = 12 7 = . So p=12*(
The total energy consumed is the sum of the energy consumed by the two power sources of 279+
A key aspect of this problem lies in the direction of the current and the positive or negative power supply. Keep in mind that the positive side of the resistor is in the opposite direction of the arrow, and you may get a negative voltage (when the current is negative), remember to keep the symbol for the calculation.
3.First, find i, i=24 (7+5)=2a, so the current source is.
According to KVL, the voltage of the current source is V=2*15+24=54V, so the power is 54*2=108W
According to the KCL, the current source does not output any current, so the power is 0
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Anonymous users2024-02-041.The current equation for the A and B node columns.
a: -i1-i2-i0=0
b: i2+i3-is=0
From i0=-2a, i1=-4a, i2=6a can be obtained. Substituting i2=6a and is=8a into b gives i3=2a
2. kirchoff's voltage law means that the sum of voltages in each circuit is 0, assuming clockwise. For the left loop, there is 3+v2-5=0, v2=2v, for the right loop, -10+v1-v2=0, substituting v2=2v, then v1=12v
3.This question is used by KVL and KCL together. First of all, according to KVL, the voltage direction of D is determined to be positive and negative, which is 10V.
Assuming that a is positive and negative, and writing kvl for the left loop column has -3+10+5+va=0, and va is therefore -12v, we can see that its actual polarity is the opposite of the assumption. From the KCL, it can be seen that the left loop current also flows clockwise and is 5A.
If the current flows in from the positive end, it consumes power, and if it flows from the negative end, it emits power, and according to the above judgment, AB transmits power, and CDE absorbs power.
aThe transmit power is 12*5=60W
b transmit power is 3 * 5 = 15W
c The absorbed power is 5*5=25W
d The absorbed power is 3*10=30W
The absorbed power is 2*10=20W
60+15-25-30-20=0, power is conserved.
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Anonymous users2024-02-03Is there a text version, or send me the original document.
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Anonymous users2024-02-021)q0=c1u0=120*8*10^(-6)c=2)w1=c1u0^2/2=120*120*8*10^(-6)/2j=
3) When the switch hits S2 and the circuit is stable.
Let the voltage at both ends of C1 be U1, the voltage at both ends of C2 be U2, the voltage at both ends of C3 be U3, and the voltage at both ends of C4 be U4, since C2=C3=C4=
Therefore there is u2=u3=u4
Then we can get 3u2=u1 from kvl, and square the two sides of this equation to get 9u2 2=u1 2 (1).
Conservation according to energy.
There is 3c2u2 2 2+c1u1 2 2=c1u0 2 2 (2).
It can be solved by equation (1) and (2).
u2=16√6v u1=48√6v
So u3 = u4 = u2 = 16 6v
4) Capacitors are energy storage components.
Conservation according to energy.
It can be seen that w2=w1=
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Anonymous users2024-02-011.(1) When the switch is pressed, the current of the branch where the resistor and capacitor are located (when the switch is pressed down, the current ignores the resistance on the main circuit and treats the capacitor as a wire.) Resistance on the trunk current = 100 20 = 5A Resistance current on the branch = 0a
Capacitance current on the branch = 5A
2) After stabilization, the capacitor is not energized, and the branch resistance is energized.
Trunk resistance Current = 100 40 =
Branch resistance current =
Trunk capacitance current = 0A
3) After stabilization, the voltage on the capacitor is the voltage of the branch resistor = * 20 = 50v
4) I don't know, I won't solve.
14 Nor will it.
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