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A battery of four 1.5 V cells in series powers a circuit with two 200 ohm resistors in parallel, connected to a 400 ohm resistor in series. What is the voltmeter reading across the 400 ohm resistor?

Physics

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Answer

4.8 V\text{4.8 V}

Step-by-step solution

4 steps

To calculate the reading on the voltmeter, we need to find the voltage across the 400 Ω400 \, \Omega resistor. This requires determining the total voltage of the battery and the total current flowing through the circuit.

Assumption: The battery consists of four cells connected in series. Assuming each cell provides 1.5 V1.5 \, \text{V}, the total voltage of the battery is 4×1.5 V=6 V4 \times 1.5 \, \text{V} = 6 \, \text{V}.

  1. Calculate the equivalent resistance of the two 200 Ω200 \, \Omega resistors connected in parallel.

    1Rp=1200 Ω+1200 Ω\frac{1}{R_{p}} = \frac{1}{200 \, \Omega} + \frac{1}{200 \, \Omega} 1Rp=2200 Ω\frac{1}{R_{p}} = \frac{2}{200 \, \Omega} Rp=200 Ω2=100 ΩR_{p} = \frac{200 \, \Omega}{2} = 100 \, \Omega
  2. Calculate the total equivalent resistance of the entire circuit.

    The parallel combination (RpR_p) is in series with the 400 Ω400 \, \Omega resistor.

    Rtotal=Rp+400 ΩR_{total} = R_p + 400 \, \Omega Rtotal=100 Ω+400 Ω=500 ΩR_{total} = 100 \, \Omega + 400 \, \Omega = 500 \, \Omega
  3. Calculate the total current flowing from the battery.

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

    Itotal=VtotalRtotalI_{total} = \frac{V_{total}}{R_{total}} Itotal=6 V500 ΩI_{total} = \frac{6 \, \text{V}}{500 \, \Omega} Itotal=0.012 AI_{total} = 0.012 \, \text{A}
  4. Calculate the voltage drop across the 400 Ω400 \, \Omega resistor.

    Since the 400 Ω400 \, \Omega resistor is in series with the parallel combination, the total current ItotalI_{total} flows through it. The voltmeter measures this voltage.

    V400=Itotal×400 ΩV_{400} = I_{total} \times 400 \, \Omega V400=0.012 A×400 ΩV_{400} = 0.012 \, \text{A} \times 400 \, \Omega V400=4.8 VV_{400} = 4.8 \, \text{V}

    The reading on the voltmeter is 4.8 V\text{4.8 V}.

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