Circuit Physics Example 3

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Example 3

medium
A circuit has a 9 V9 \text{ V} battery and a 15 Ω15 \text{ } \Omega resistor. Find the current flowing and the power consumed.

Solution

  1. 1
    Apply Ohm's law to find the current: I=VR=915=0.6 AI = \frac{V}{R} = \frac{9}{15} = 0.6 \text{ A}.
  2. 2
    Calculate the power using P=VIP = VI: P=9×0.6=5.4 WP = 9 \times 0.6 = 5.4 \text{ W}
  3. 3
    Verify with P=I2R=(0.6)2×15=0.36×15=5.4 WP = I^2 R = (0.6)^2 \times 15 = 0.36 \times 15 = 5.4 \text{ W}. Consistent.

Answer

I=0.6 A,P=5.4 WI = 0.6 \text{ A}, \quad P = 5.4 \text{ W}
A simple circuit with one battery and one resistor is the most fundamental circuit. The current is determined by Ohm's law, and the power consumed by the resistor equals the power delivered by the battery.

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