A.)For the real battery shown in the figure, calculate the power dissipated by a

Question

A.)For the real battery shown in the figure, calculate the power dissipated by a resistor R connected to the battery when R = 2.3? .

B.)For the real battery shown in the figure, calculate the power dissipated by a resistor R connected to the battery when R = 5.0? .

C.)For the real battery shown in the figure, calculate the power dissipated by a resistor R connected to the battery when R = 10? .

D.)For the real battery shown in the figure, calculate the power dissipated by a resistor R connected to the battery when R = 20? . (Your results should suggest that maximum power dissipation is achieved when the external resistance R equals the internal resistance. This is true in general.)

E.)For the real battery shown in the figure, calculate the power dissipated by a resistor R connected to the battery when R = 43?

Explanation / Answer

Power dessipated by R is given by:- P=i^2 *R where i=current flowing through R.

Part A:- Current obtained = 1.5/(10+2.3)= 0.121 Ampere
So,Power dessipated by R= 0.034 Watt.

Part B:- Current obtained = 1.5/(10+5)=0.1 Ampere
So,Power dessipated by R= 0.050 Watt.

Part C:- Current obtained = 1.5/(10+10)=0.075 Ampere
So,Power dessipated by R= 0.056 Watt.

Part D:- Current obtained = 1.5/(10+20)=0.05 Ampere
So,Power dessipated by R= 0.050 Watt.

Part E:- Current obtained = 1.5/(10+43)=0.028 Ampere
So,Power dessipated by R= 0.034 Watt.

So, we can see that Power dessipated in part C is maximum, when external resistance R equals the internal resistance.

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