A 12-voltage battery is connected to a light bulb and draws 150mA of current. What is the resistance, power, and how much to operate it for a month if electricity is K11 per kWh?

To calculate the resistance of the light bulb, we can use Ohm's Law:

Resistance (R) = Voltage (V) / Current (I)
R = 12V / 0.15A
R = 80 ohms

Next, we can calculate the power consumed by the light bulb using the formula:

Power (P) = Voltage (V) x Current (I)
P = 12V x 0.15A
P = 1.8 watts

To calculate the energy consumed by the light bulb in one hour, we can use the formula:

Energy (E) = Power (P) x Time (T)
E = 1.8W x 1 hour
E = 1.8 watt-hours

To calculate the energy consumed by the light bulb in one month (30 days), we can use the formula:

Energy (E) = Power (P) x Time (T)
E = 1.8W x 24 hours x 30 days
E = 1296 watt-hours

To convert the energy consumed to kilowatt-hours (kWh), we divide by 1000:

Energy (E) = 1296 watt-hours / 1000
E = 1.296 kWh

Finally, to calculate the cost of operating the light bulb for a month, we multiply the energy consumed by the cost of electricity:

Cost = Energy (E) x Cost per kWh
Cost = 1.296 kWh x K11
Cost = K14.26

Therefore, the resistance of the light bulb is 80 ohms, the power consumed is 1.8 watts, and it will cost K14.26 to operate it for a month.

Resistance:

Resistance (R) = Voltage (V) / Current (I)
R = 12 volts / 0.150 amps
R = 80 ohms

Power:

Power (P) = Voltage (V) x Current (I)
P = 12 volts x 0.150 amps
P = 1.8 watts

Cost to Operate for a Month:

Assuming the bulb is on for 24 hours a day, for 30 days in a month:

Energy consumption (E) = Power (P) x Time (t)
E = 1.8 watts x 24 hours/day x 30 days
E = 1296 watt-hours (Wh) = 1.296 kilowatt-hours (kWh)

Cost = Energy consumption (kWh) x Electricity cost (Ksh/kWh)
Cost = 1.296 kWh x Ksh 11/kWh
Cost = Ksh 14.26

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