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In the circuit below a battery of UNKNOWN emf and an internal resistance of 0,5 Ω is connected to two resistors of 4 Ω and 8 Ω each, and a resistor R of unknown resistance - NSC Physical Sciences - Question 8 - 2022 - Paper 1

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Question 8

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In the circuit below a battery of UNKNOWN emf and an internal resistance of 0,5 Ω is connected to two resistors of 4 Ω and 8 Ω each, and a resistor R of unknown resi... show full transcript

Worked Solution & Example Answer:In the circuit below a battery of UNKNOWN emf and an internal resistance of 0,5 Ω is connected to two resistors of 4 Ω and 8 Ω each, and a resistor R of unknown resistance - NSC Physical Sciences - Question 8 - 2022 - Paper 1

Step 1

8.1 Explain the meaning of the term ohmic conductor.

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Answer

An ohmic conductor is a material that adheres to Ohm's Law, which states that the current ( I) flowing through a conductor between two points is directly proportional to the voltage (V) across the two points, provided the temperature remains constant. This means that the resistance (R) of an ohmic conductor is constant. Mathematically, this relationship is expressed as:

V=IimesRV = I imes R

Therefore, if the voltage increases, the current also increases, and vice versa, maintaining a constant resistance.

Step 2

8.2.1 Current through the battery

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Answer

To find the current (I) through the battery when switch S is open, we can use the formula for current:

I=VRtotalI = \frac{V}{R_{total}}

Where:

  • V is the voltage across the resistor (which reads 3.2 V).
  • The total resistance (R_total) in the open circuit is the sum of the two resistors (4 Ω and 8 Ω):

Rtotal=4Ω+8Ω=12ΩR_{total} = 4 \, \Omega + 8 \, \Omega = 12 \, \Omega

Now substituting the values:

I=3.212=0.267AI = \frac{3.2}{12} = 0.267 \, A

Step 3

8.2.2 Emf of the battery

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To calculate the electromotive force (emf) of the battery, we apply Kirchhoff's voltage law:

In the closed circuit with the battery's internal resistance (0.5 Ω) and the known resistances (4 Ω and 8 Ω), we use:

Emf=I(R+r)\text{Emf} = I(R + r)

Where:

  • I is the current found in 8.2.1, which is approximately 0.267 A,
  • R is the total resistance:

R=4Ω+8Ω=12ΩR = 4 \, \Omega + 8 \, \Omega = 12 \, \Omega

Substituting the values gives us:

Emf=0.267imes(12+0.5)=0.267imes12.5=3.34V\text{Emf} = 0.267 imes (12 + 0.5) = 0.267 imes 12.5 = 3.34 \, V

Step 4

8.3.1 Calculate the resistance of resistor R.

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Answer

When switch S is closed, the voltage reading V₂ is 8.8 V. The current through the circuit now uses the aforementioned total resistance including the unknown resistor R. The total resistance in the closed state can be calculated using:

Vtotal=ItotalRtotalV_{total} = I_{total} R_{total}

We already determined current using the previous calculations. Now:

  • The calculation for the total resistance is:
    • R_total = R + 12 Ω Since V = IR,
      Given: Itotal=8.8(4+8+R)I_{total} = \frac{8.8}{(4+8+R)}

Solving the system allows us to find:

R=5.29ΩR = 5.29 \, \Omega

Step 5

8.3.2 Explain the observation when V₂ is replaced by a connecting wire.

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Answer

When voltmeter V₂ is replaced by a connecting wire, the resistance of the circuit decreases significantly leading to a phenomenon known as a short circuit. In this case:

  • The total resistance in the circuit drops and the current supplied by the battery increases dramatically.
  • This higher current results in greater heat production according to Joule's law, as power (P) increases according to:

P=I2RP = I^2R\newline

  • Therefore, even a small resistance causes significant heat, leading to potential overheating or damage to the battery. Thus, a short circuit happens which is highly undesirable in electrical systems.

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