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Calculate the equivalent resistance of the parallel combination of the 4 and 6 resistors.

Physics

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Part a· 2 steps

The combined resistance of the above circuit.

  1. Calculate the equivalent resistance of the parallel combination of the 4 Ω4\,\Omega and 6 Ω6\,\Omega resistors.

    1Rp=14 Ω+16 Ω\frac{1}{R_p} = \frac{1}{4\,\Omega} + \frac{1}{6\,\Omega} 1Rp=312 Ω+212 Ω\frac{1}{R_p} = \frac{3}{12\,\Omega} + \frac{2}{12\,\Omega} 1Rp=512 Ω\frac{1}{R_p} = \frac{5}{12\,\Omega} Rp=125 Ω=2.4 ΩR_p = \frac{12}{5}\,\Omega = 2.4\,\Omega
  2. Calculate the total combined resistance (RtotalR_{total}) by adding the series resistor (2 Ω2\,\Omega) to the parallel equivalent resistance (RpR_p).

    Rtotal=2 Ω+RpR_{total} = 2\,\Omega + R_p Rtotal=2 Ω+2.4 ΩR_{total} = 2\,\Omega + 2.4\,\Omega Rtotal=4.4 ΩR_{total} = 4.4\,\Omega

    The combined resistance of the circuit is 4.4 \Omega\boxed{\text{4.4 \Omega}}.

Part b· 1 step

The current flowing through the ammeter.

  1. Use Ohm's Law (V=I⋅RV = I \cdot R) to find the total current (ItotalI_{total}) flowing from the voltage source, which is measured by the ammeter.

    Itotal=VsourceRtotalI_{total} = \frac{V_{source}}{R_{total}} Itotal=12 V4.4 ΩI_{total} = \frac{12\,V}{4.4\,\Omega} Itotal=12044 A=3011 AI_{total} = \frac{120}{44}\,A = \frac{30}{11}\,A Itotal≈2.73 AI_{total} \approx 2.73\,A

    The current flowing through the ammeter is 2.73 A\boxed{\text{2.73 A}}.

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