When the power supply voltage remains unchanged and the resistance of a resistor increases by 3 ohm, the current intensity becomes four fifths of the original?

When the power supply voltage remains unchanged and the resistance of a resistor increases by 3 ohm, the current intensity becomes four fifths of the original?

If the original resistance is r, the voltage is u, and the current is I, then I = ur ------ (1) 45i = ur + 3 Ω ------ (2) substitute (1) into (2), and the solution is r = 12 Ω. Answer: the original resistance is 12 Ω

In a 15V series circuit, there is a resistance of 5 ohm and a lamp of 10 power. What is the resistance of the bulb and the rated voltage of the bulb

I*I*R+W=U*I
I*I*5+10=15I
Finishing: I * i-3i + 2 = 0
I = 1 or I = 2
When I = 1
R = w / (I * I) = 10 / 1 = 10 Ω voltage U = 10V
When I = 2
R = w / (I * I) = 10 / 4 = 2.5 Ω, voltage U = 5V

In the process, there is a bulb with 484 ohm resistance connected to the home circuit. How much power is it? How much current is it? If the voltage rises by 10% during the day
There is a bulb with 484 ohm resistance connected to the home circuit. How much power is it? How much current is it? If the voltage rises by 10% during the day and falls by 10% at night, what is the actual power of the bulb?

Analysis:
1. According to the formula P = UI
R = u / I I = u / R
Substituting I = u / R to calculate P = u * (U / R)
2. Since the household electricity is 220 V and 50 Hz, the input voltage can be obtained
3. When the daytime voltage rises by 10%, the input voltage is 220 + 220 * 10% = 242v
therefore
P=242*(242/484)=121W
4. Because the voltage drops by 10% at night, the input voltage is 220-220 * 10% = 198v
therefore
P=198*(198/484)=81W

A small bulb with rated voltage less than 6V and rated power of 1W. When it is connected to the circuit in parallel with an 18 ohm resistor
If the current in the main road is 0.5A and the bulb can light normally, the rated voltage of the small bulb is____ 5. The resistance of the small bulb is____ Ohm, the current through the bulb is____ A.

If the rated voltage of small bulb is u, its resistance R = u * U / P = u * U / 1 = u * u ohm,
When the small bulb and resistor R1 are connected in parallel and light up normally, the power supply voltage should be u,
U/R + U/R1 = 0.5 1/U + U/18 = 0.5 U*U- 9*U +18=0 （U-6）* （U-3）=0
Since u is less than 6 V, u = 3 V
R = u * u = 3 * 3 = 9 ohm
I=U/R=3/9=0.3333 A