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A circuit is set up as shown - Scottish Highers Physics - Question 15 - 2018

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A circuit is set up as shown. The battery has negligible internal resistance. A student makes the following statements about the readings on the meters in this cir... show full transcript

Worked Solution & Example Answer:A circuit is set up as shown - Scottish Highers Physics - Question 15 - 2018

Step 1

I When switch S is open the reading on the voltmeter will be 6.0 V.

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Answer

When switch S is open, the entire 12V from the battery is seen across the 10Ω resistor in series with the voltmeter. Thus, the voltmeter will read 6.0V since the total voltage drop is divided equally across two 10Ω resistors in parallel.

Step 2

II When switch S is open the reading on A₁ will be 0.60 A.

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Answer

Using Ohm's law, we can calculate the total resistance in the circuit. The two 10Ω resistors in parallel provide a total resistance of 5Ω. The current can be calculated using the formula:

I=VR=12V5Ω=2.4AI = \frac{V}{R} = \frac{12V}{5Ω} = 2.4 A

As A₁ reads the current through just one of the resistors, the current through A₁ becomes ( 2.4 A / 4 = 0.60 A ).

Step 3

III When switch S is closed the reading on A₁ will be 0.80 A.

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Answer

With switch S closed, the total circuit resistance decreases. The current I can again be calculated as follows, recognizing that two 10Ω resistors are in parallel, giving us a total equivalent resistance of 5Ω. So,

I=12V4Ω=3AI = \frac{12V}{4 Ω} = 3 A

The current divides evenly among parallel paths, leading to each path carrying ( \frac{3 A}{4} = 0.80 A ) through A₁.

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