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A 4 ohm resistor is connected in series to a battery of e.m.f. 6 V and negligible internal resistance. Determine the power dissipated by the resistor.

      

A 4 ohm resistor is connected in series to a battery of e.m.f. 6 V and negligible internal resistance. Determine the power dissipated by the resistor.

  

Answers


John
P = $\frac{V^2}{R}$

= $\frac{6\times 6}{4}$

= 9 Watts
johnmulu answered the question on February 15, 2017 at 13:21


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