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Will the current increase or decrease, if two more cells are added to the circuit in 7.4? Check your prediction using the formula.

Physics

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Answer

The current will increase to 3 A\text{The current will increase to } 3 \text{ A}

Step-by-step solution

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Okay, Success💗🎀🥹, let's tackle question 7.5.

First, we need to determine the resistance of the resistor from question 7.4, as it will be the same resistor in question 7.5.

From question 7.4:

  • Voltage (VV) = 6 V6 \text{ V}
  • Current (II) = 2 A2 \text{ A}
  1. Calculate the resistance (R1R_1) using Ohm's Law (R=VIR = \frac{V}{I}).

    R1=6 V2 AR_1 = \frac{6 \text{ V}}{2 \text{ A}} R1=3 ΩR_1 = 3 \ \Omega

    The resistance of the resistor is 3 Ω3 \ \Omega.

    For question 7.5:
    The question states that two more cells are added to the circuit from 7.4.
    An original 6 V battery typically consists of 4 cells (assuming 1.5 V per cell). Adding two more cells means adding 2×1.5 V=3 V2 \times 1.5 \text{ V} = 3 \text{ V} to the original voltage.
    So, the new voltage is 6 V+3 V=9 V6 \text{ V} + 3 \text{ V} = 9 \text{ V}. This matches the 9 V battery shown in the diagram for 7.5.
    The resistance (R1R_1) remains the same at 3 Ω3 \ \Omega.

    Prediction: If the voltage increases and the resistance stays the same, the current will increase.

  2. Identify the new voltage and the resistance for the circuit in 7.5.

    New Voltage (V1V_1) = 9 V9 \text{ V}
    Resistance (R1R_1) = 3 Ω3 \ \Omega

  3. Calculate the new current (I1I_1) using Ohm's Law (I=VRI = \frac{V}{R}).

    I1=V1R1I_1 = \frac{V_1}{R_1} I1=9 V3 ΩI_1 = \frac{9 \text{ V}}{3 \ \Omega} I1=3 AI_1 = 3 \text{ A}
  4. Compare the new current with the original current.

    Original current = 2 A2 \text{ A}
    New current = 3 A3 \text{ A}
    Since 3 A>2 A3 \text{ A} > 2 \text{ A}, the current will increase.

    The current will increase to 3 A\text{The current will increase to } 3 \text{ A}
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