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Three resistors are connected to a 3.0 V power supply as shown - Scottish Highers Physics - Question 19 - 2022

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Three resistors are connected to a 3.0 V power supply as shown. The power supply has negligible internal resistance. The power dissipated in the circuit is: A. 0.25... show full transcript

Worked Solution & Example Answer:Three resistors are connected to a 3.0 V power supply as shown - Scottish Highers Physics - Question 19 - 2022

Step 1

Calculate Total Resistance

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Answer

In the circuit, the 6.0 Ω resistors are in parallel. The equivalent resistance can be calculated using the formula:

rac{1}{R_{eq}} = rac{1}{R_1} + rac{1}{R_2}

Substituting the values:

rac{1}{R_{eq}} = rac{1}{6} + rac{1}{6} = rac{2}{6} = rac{1}{3} ightarrow R_{eq} = 3.0 \, \Omega

Now, adding the resistance of the 9.0 Ω resistor:

Rtotal=Req+R3=3.0 Ω+9.0 Ω=12.0 ΩR_{total} = R_{eq} + R_3 = 3.0 \, \Omega + 9.0 \, \Omega = 12.0 \, \Omega

Step 2

Calculate Current in the Circuit

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Answer

Using Ohm's Law, the total current (I) can be calculated using:

I=VRtotal=3.0 V12.0 Ω=0.25 AI = \frac{V}{R_{total}} = \frac{3.0 \, V}{12.0 \, \Omega} = 0.25 \, A

Step 3

Calculate Power Dissipated in the Circuit

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Answer

The power dissipated (P) can be calculated using the formula:

P=I2⋅Req=(0.25 A)2⋅3.0 Ω=0.0625 A2⋅3.0 Ω=0.1875 WP = I^2 \cdot R_{eq} = (0.25 \, A)^2 \cdot 3.0 \, \Omega = 0.0625 \, A^2 \cdot 3.0 \, \Omega = 0.1875 \, W

However, the power dissipated across all the resistors can also be calculated using:

P=V⋅I=3.0 V⋅0.25 A=0.75 WP = V \cdot I = 3.0 \, V \cdot 0.25 \, A = 0.75 \, W

Thus, the power dissipated in the circuit is 0.75 W.

Step 4

Select the Correct Answer

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Answer

Based on the calculation, the answer is C. 0.75 W.

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