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Calculate the total resistance and voltage for a series circuit with R1=15Ω, R2=30Ω, R3=45Ω and I=1.5A. Also, find the energy consumed by a 100W bulb used for 8 hours.

Physics

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Answers

2 parts
a

0.8 kWh\text{0.8 kWh}

b

1000 W\text{1000 W}

Step-by-step solution

4 steps

Here are the solutions to the problems.

** (Circuit with three resistors in series):**

Step 1: Calculate the total resistance of the series circuit.
Given resistors: R1=15 ΩR_1 = 15 \, \Omega, R2=30 ΩR_2 = 30 \, \Omega, R3=45 ΩR_3 = 45 \, \Omega.
For resistors connected in series, the total resistance (RtotalR_{total}) is the sum of the individual resistances:

Rtotal=R1+R2+R3R_{total} = R_1 + R_2 + R_3 Rtotal=15 Ω+30 Ω+45 ΩR_{total} = 15 \, \Omega + 30 \, \Omega + 45 \, \Omega Rtotal=90 ΩR_{total} = 90 \, \Omega

Step 2: Calculate the total voltage applied to the circuit using Ohm's Law.
Given total current (II) = 1.5 A1.5 \, \text{A}.
According to Ohm's Law, the total voltage (VtotalV_{total}) is given by:

Vtotal=I×RtotalV_{total} = I \times R_{total} Vtotal=(1.5 A)×(90 Ω)V_{total} = (1.5 \, \text{A}) \times (90 \, \Omega) Vtotal=135 VV_{total} = 135 \, \text{V}

The total voltage applied to the circuit is 135 V\boxed{\text{135 V}}.

** (Electrical energy and power rating):**

Part a· 2 steps

How much electrical energy is consumed by the bulb in a day?

  1. Identify the given power and time.

    Power (PP) of the light bulb = 100 W100 \, \text{W}.
    Time (tt) of use per day = 8 hours8 \, \text{hours}.

  2. Calculate the energy consumed.

    Energy (EE) is calculated as Power ×\times Time. To get the energy in kilowatt-hours (kWh), convert power from Watts to kilowatts.

    P=100 W=1001000 kW=0.1 kWP = 100 \, \text{W} = \frac{100}{1000} \, \text{kW} = 0.1 \, \text{kW} E=P×tE = P \times t E=(0.1 kW)×(8 hours)E = (0.1 \, \text{kW}) \times (8 \, \text{hours}) E=0.8 kWhE = 0.8 \, \text{kWh}

    The electrical energy consumed by the bulb in a day is 0.8 kWh\boxed{\text{0.8 kWh}}.

Part b

1000 W\text{1000 W}

An electric heater consumes 2 kWh of electrical energy in 2 hours. What is its power rating?

  1. Identify the given energy and time.

    Energy (EE) consumed = 2 kWh2 \, \text{kWh}.
    Time (tt) taken = 2 hours2 \, \text{hours}.

  2. Calculate the power rating.

    Power (PP) is calculated as Energy divided by Time.

    P=EtP = \frac{E}{t} P=2 kWh2 hoursP = \frac{2 \, \text{kWh}}{2 \, \text{hours}} P=1 kWP = 1 \, \text{kW}

    To express in Watts:

    P=1 kW=1000 WP = 1 \, \text{kW} = 1000 \, \text{W}

    The power rating of the electric heater is 1000 W\boxed{\text{1000 W}}.

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