Buzz
Solve problem C1 on page 40 of Physics 9 textbook.
Problem Statement:
When connecting a light bulb to a voltage of 220V, the current flowing through it is 341mA.
a) Calculate the electrical resistance and power of the light bulb at that moment.
b) This light bulb is used in the same conditions, averaging 4 hours a day. Calculate the electrical energy consumed by the light bulb in 30 days in joules and the corresponding electricity meter reading.
HINT FOR SOLUTION:
a) Calculate the electrical resistance R of the light bulb.
Calculate the power consumption P of the light bulb.
b) Calculate the electrical energy A consumed by the light bulb.
Calculate the meter reading N of the electricity meter.
Solution:
a)
The power of the light bulb is: P = UI = 220.0,341 = 75W.
b) The electrical energy consumed by the light bulb in 30 days, 4 hours each day is:
The electrical energy A consumed by the light bulb is: A = Pt = 75.30.4.3600 = 32400000 (J)
Each count of the electricity meter corresponds to 1 kWh, so to find the meter reading, we need to calculate the electrical energy in kWh.
In that case, A = Pt = 75.30.4 = 9000Wh = 9kWh
Therefore, the meter reading of the electricity meter is 9 digits.
Solution for problem C2 on page 40 of Physics 9 textbook
Problem Statement:
A circuit consists of a light bulb labeled 6V - 4.5W connected in series with a variable resistor, placed in a constant 9V voltage source as shown in figure 14.1. The resistance of the connecting wire and the ammeter is negligible.
a) Close switch K, the light bulb lights up normally. Calculate the reading on the ammeter.
Solution for problem C2 on page 40 of Physics 9 textbook
Problem Statement:
A circuit consists of a light bulb labeled 6V - 4.5W connected in series with a variable resistor, placed in a constant 9V voltage source as shown in figure 14.1. The resistance of the connecting wire and the ammeter is negligible.
a) When switch K is closed, the light bulb lights up normally, indicating that the current intensity through the bulb is equal to the rated current, and that is also the reading on the ammeter.
Calculations for problem C2 on page 40 of Physics 9 textbook
Results:
a) The voltage drop across the variable resistor is: 9V - 6V = 3V.
b) The resistance Rbt of the variable resistor is: Rbt = Vbt / I = 3V / 0.341A = 8.79Ω.
Solution:
a) When switch K is closed, the light bulb lights up normally, indicating that the current intensity through the bulb is equal to the rated current, and that is also the reading on the ammeter.
Solve problem C3 on page 41 of Physics 9 textbook
Problem Statement:
A hairpin light bulb marked 220V - 100W and a table marked 220V - 1000W are both connected to a 220V power outlet at home to operate normally.
a) Draw the circuit diagram, where the table is symbolized as a resistor, and calculate the equivalent resistance of this circuit.
b) Calculate the electrical energy consumed by this circuit in 1 hour in both joules and kilowatt-hours.
SUGGESTED SOLUTION
a) Illustrate the electrical circuit diagram.
Calculate the resistance of the light bulb R1 = 484 Ω.
Determine the resistance of the table R2 = 48.4 Ω.
Find the equivalent resistance R of the circuit segment.
b) Calculate the electrical energy A consumed by the circuit in 1 hour.
Solution:
a) For the lamp and the desk to operate normally when connected to a voltage of 220V, they must be connected in parallel. Here is the circuit diagram:
Chapter I Electrical Science - Exercise applying Jun-Lenz laws. Refer to the solution hints on pages 47, 48 Physics 9 for effective Physics 12 study.
Jun-Lenz Law is the next topic in Chapter I of Electrical Science Physics 9 Grade 11. Also, refer to the solution hints on page 45 Physics 9 to grasp the knowledge and excel in Physics 9.
