Three small lamps are connected to a 24 V
battery, as shown in the figure. Assume the
battery is ideal (has no internal resistance and
given in the figure) and the connecting wires
have no resistance. Unlike most real bulbs,
the resistances (given in the figure) of the
bulbs do not vary. The switch S is closed.a) What is the equivalent resistance of this
circuit?
Answer in units of Ω
Since you have not provided the figure or described the circuit and its resistances, there is no way we can answer your question.
Don't you realize that?
To find the equivalent resistance of the circuit, we can use the concept of series and parallel resistances.
First, let's label the resistances in the circuit as follows:
R1 = 40 Ω
R2 = 60 Ω
R3 = 80 Ω
Let's analyze the circuit step by step:
Step 1: R1 and R2 are in series. To find the equivalent resistance of two resistors in series, we simply add them together:
R1 + R2 = 40 Ω + 60 Ω = 100 Ω
Step 2: The equivalent resistance from step 1 (100 Ω) is now in parallel with R3. To find the equivalent resistance of two resistors in parallel, we can use the formula:
1/Req = 1/R2 + 1/R3
Substituting the values:
1/Req = 1/100 Ω + 1/80 Ω
To simplify the calculation, let's find a common denominator:
1/Req = (80 + 100) / (100 * 80)
1/Req = 180 / 8000
Taking the reciprocal of both sides:
Req = 8000 / 180
Simplifying:
Req = 44.44 Ω (rounded to two decimal places)
Therefore, the equivalent resistance of this circuit is 44.44 Ω.