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The small bulb L marked with "6V & nbsp; & nbsp; 3W" and the resistor R with rated power of 4W are connected according to the circuit shown in the figure. When the switch S is closed, the lamp l will light normally, and the current indication is 0.8A. Calculate: (1) the resistance value of the resistor R; (2) the total power consumed by the small bulb and the resistor at this time
(1) Lamp L lights normally, the current through bulb IL = Pu = 3w6v = 0.5A, the current through resistor I1 = i-il = 0.8a-0.5a = 0.3A, the resistance value of resistor R = ui1 = 6v0.3a = 20 Ω. Answer: the resistance value of resistor R is 20 Ω. (2) at this time, the total power consumed by small bulb and resistor P = UI = 6V × 0.8A = 4.8w
The power supply voltage is 8V, the resistance value of resistor R1 is 20 Ω, the resistance value of bulb is 12 Ω, the bulb specification is "6V 3W", and the total power in the circuit
R1r2 series connection
Total resistance R = 20 + 12 = 32 Ω
Total current I = u / r = 8 / 32 = 1 / 4A
Total power P = UI = 8 * 1 / 4 = 2W
In the circuit as shown in the figure, the rated power of lamp L is 2W, R2 = 10 Ω, and the resistance of small bulb R1
1
When S1 and S2 are disconnected, the circuit is in series. Only R2 and lamp l work in the whole circuit
P lamp amount = u lamp amount I lamp amount
So I lamp amount = P lamp amount / u lamp amount
Because U2 = i2r2
So I2 = U2 / R2
Because I = I light = I2
So U2 / r2 = P / u (equation 1)
U = U1 + U2 in series circuit
So U2 + U light amount = 9 (equation 2)
The two formulas are solved simultaneously
It's important to take advantage of this relationship
U = 4V or 5V
Based on the principle of voltage division in series circuit, R2 > R lamp
So u = 4V
U2=5V
I=U2/R2=5V/10Ω ＝0.5A
R1 = u light level / I = 4V / 0.5A = 8 Ω
2
S1, S2 are closed, the bulb short circuit. The circuit is in parallel, only R2, R3 work. U = U1 = U2 = U3
P total = UI total
So I total = P total / u = 24.3w/9v = 2.7A
I2=U2/R2=9V/10Ω ＝0.9A
I3 = I total - I2 = 2.7a-0.9a = 1.8A
R3=U3/I3=9V/1.8A=5Ω
If the bulb marked "6V & nbsp; 6W" is connected to the circuit with voltage of 15V and makes it light normally, it should______ Connect a resistance value of______ The resistance is less than one ohm
∵ P = UI, rated current of the bulb: I = Pu = 6w6v = 1a, in order to make the bulb shine normally, the voltage at both ends of the bulb should be 6V, the current in the circuit is I = 1a, the voltage of series resistance should be u ′ = 15v-6v = 9V, according to Ohm's law, the resistance of series resistance can be obtained: r = u ′, I = 9v1a = 9 Ω; therefore, the answer is: series, 9
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