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I) A circuit diagram for two lamps (resistors) connected in parallel to a battery:

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2 parts
c(ii)

1.5 Ω\text{1.5 } \Omega

c(iii)

3 A\text{3 A}

Step-by-step solution

4 steps

Here's a detailed breakdown for part c):

Part c(i)

A circuit diagram for two lamps (resistors) connected in parallel to a battery

graph TD
    A[Battery +] --- B
    B --- C{Lamp 1}
    B --- D{Lamp 2}
    C --- E
    D --- E
    E --- F[Battery -]

This diagram represents a battery connected to two parallel branches. Each branch contains a lamp. The standard symbols are:

  • Battery: A long line and a short line, representing the positive and negative terminals respectively.
  • Lamp/Resistor: A circle with a cross inside, or a zigzag line.

Part c(ii)

1.5 Ω\text{1.5 } \Omega

Calculate the effective resistance of the lamps.

  1. Identify the resistance of each lamp and the connection type.

    Each lamp has a resistance R=3 ΩR = 3 \, \Omega. They are connected in parallel.

  2. Use the formula for effective resistance in a parallel circuit.

    For two resistors in parallel:

    1Reff=1R1+1R2\frac{1}{R_{eff}} = \frac{1}{R_1} + \frac{1}{R_2}

    Substitute the given values:

    1Reff=13 Ω+13 Ω\frac{1}{R_{eff}} = \frac{1}{3 \, \Omega} + \frac{1}{3 \, \Omega} 1Reff=23 Ω\frac{1}{R_{eff}} = \frac{2}{3 \, \Omega}

    To find ReffR_{eff}, take the reciprocal of both sides:

    Reff=32 ΩR_{eff} = \frac{3}{2} \, \Omega Reff=1.5 ΩR_{eff} = 1.5 \, \Omega

    The effective resistance is 1.5 Ω\boxed{\text{1.5 } \Omega}.

Part c(iii)

3 A\text{3 A}

Calculate the total current in the circuit.

  1. Identify the total voltage and effective resistance.

    The battery voltage V=4.5 VV = 4.5 \, V.
    The effective resistance Reff=1.5 ΩR_{eff} = 1.5 \, \Omega (calculated in part ii).

  2. Use Ohm's Law to calculate the total current.

    Ohm's Law states I=VRI = \frac{V}{R}.

    I=4.5 V1.5 ΩI = \frac{4.5 \, V}{1.5 \, \Omega} I=3 AI = 3 \, A

    The total current in the circuit is 3 A\boxed{\text{3 A}}.

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