Explanation Given data: A battery and a resistor are wired into a circuit. The resistor dissipates the power is 0.50 W. Two batteries, each identical to the original one, are connected in series to the resistor. It is required to find the power dissipated by the resistor. Consider power dissipated by the resistor when it is connected to the single battery as case 1 and the power dissipated by the resistor when it is connected in series with the two identical batteries as case 2. Power dissipated by the resistor in case 1 ( P 1 ) is 0.50 W. It is required to find the power dissipated by the resistor in case 2. Consider the potential difference of each battery as Δ V . Consider the resistance of the resistor as R . Formula used: Write the expression for power dissipated by the resistor as follows. P = ( Δ V ) 2 R (1) Here, Δ V is the potential difference of the battery, and R is the resistance of the resistor. Explanation: Calculation of power dissipated by the resistor in case 2 ( P 2 ) : Modify the expression in equation (1) for the power dissipated by the resistor in case 1 as follows. P 1 = ( Δ V 1 ) 2 R Substitute 0.50 W for P 1 , and Δ V for Δ V 1 , 0.50 W = ( Δ V ) 2 R ( Δ V ) 2 R = 0

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ISBN: 8220106755235

Author: Field

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Chapter 22, Problem 17CQ

To determine

To find: The power dissipated by the resistor when the two identical batteries are connected in series.

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Chapter 22, Problem 16CQ

Chapter 22, Problem 18CQ

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