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State Ohm's law - Leaving Cert Physics - Question b - 2010

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State Ohm's law. The diagram shows a number of resistors connected to a 12 V battery and a bulb whose resistance is 4 Ω. Calculate: (i) the combined resistance of... show full transcript

Worked Solution & Example Answer:State Ohm's law - Leaving Cert Physics - Question b - 2010

Step 1

(i) the combined resistance of the 15 Ω and 30 Ω resistors in parallel.

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Answer

To find the combined resistance ( R_{combined} ) of resistors in parallel, we use the formula: 1Rcombined=1R1+1R2\frac{1}{R_{combined}} = \frac{1}{R_1} + \frac{1}{R_2}

Substituting the values: 1Rcombined=115Ω+130Ω\frac{1}{R_{combined}} = \frac{1}{15 \Omega} + \frac{1}{30 \Omega}

Calculating the right side: 1Rcombined=2+130=330\frac{1}{R_{combined}} = \frac{2 + 1}{30} = \frac{3}{30}

Thus, Rcombined=303=10ΩR_{combined} = \frac{30}{3} = 10 \Omega

Step 2

(ii) the total resistance of the circuit.

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Answer

The total resistance in the circuit can be calculated by adding the resistance of the combined parallel resistors and the resistance of the bulb:

First, we know:

  • Combined resistance of 15 Ω and 30 Ω resistors = 10 Ω.
  • Resistance of the bulb = 4 Ω.

Thus, Rtotal=Rcombined+Rbulb=10Ω+4Ω=14ΩR_{total} = R_{combined} + R_{bulb} = 10 \Omega + 4 \Omega = 14 \Omega

Step 3

(iii) the current flowing in the bulb.

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Answer

To find the current flowing in the bulb ( I ), we use Ohm's Law: V=I⋅RV = I \cdot R where:

  • Voltage ( V ) = 12 V
  • Total resistance ( R_{total} ) = 14 Ω.

Rearranging for current: I=VRtotal=12V14Ω≈0.857AI = \frac{V}{R_{total}} = \frac{12 V}{14 \Omega} ≈ 0.857 A

Therefore, the current flowing in the bulb is approximately 0.86 A.

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