Potential Difference Physics Example 3

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Example 3

medium
A 12 V12 \text{ V} battery drives 3 A3 \text{ A} of current through a resistor. Find the resistance and the power dissipated.

Solution

  1. 1
    Apply Ohm's law to find the resistance: R=VI=123=4 ΩR = \frac{V}{I} = \frac{12}{3} = 4 \text{ } \Omega.
  2. 2
    Calculate the power dissipated using P=VIP = VI: P=12×3=36 WP = 12 \times 3 = 36 \text{ W}
  3. 3
    Verify using the alternative formula: P=I2R=32×4=9×4=36 WP = I^2 R = 3^2 \times 4 = 9 \times 4 = 36 \text{ W}. Consistent.

Answer

R=4 Ω,P=36 WR = 4 \text{ } \Omega, \quad P = 36 \text{ W}
Ohm's law (V=IRV = IR) and the power formula (P=VIP = VI) are the two most fundamental relationships in circuit analysis. Power can equivalently be expressed as P=I2RP = I^2R or P=V2RP = \frac{V^2}{R}.

About Potential Difference

The difference in electric potential between two points, equal to the work done per unit charge moving between them.

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