Chapter 28: Q. 51 (page 792)

A. Load resistor R is attached to a battery of emf E and internal resistance r. For what value of the resistance R, in terms of E and r, will the power dissipated by the load resistor be a maximum?
B. What is the maximum power that the load can dissipate if the battery has $E = 9.0 V$ and localid="1649415601268" $r = 1.0 Ω$ ?

Short Answer

Expert verified

A. When R=r then power dissipation will be maximum.

B. The maximum power is $20.25 W .$

Step by step solution

Part (A) Step 1: Given information

We have given,

Emf of the battery = E

Internal resistance = r

Load resistance =R

We have to find the for value of R the power dissipation will be maximum.

Simplify

total resistance of the circuit ,

$R_{eq} = R + r$

Then current will be,

$I = \frac{E}{R_{eq}} = \frac{E}{R + r}$

Then we can write the power dissipation in the circuit will be,

localid="1649415474924" $P = {RI}^{2} = \frac{{RE}^{2}}{{(r + R)}^{2}}$

Since we have to find the value of R for maximum value of power.

Then let us maximize it,

localid="1649415432782" $\frac{d P}{d R} = 0 E^{2} [\frac{{(r + R)}^{2} - 2 R (R + r)}{{(r + R)}^{4}}] = 0 r = R$

Part (B) Step 1: Given information

We have given,

$Emf = 9 V Internal resistance = 1 Ω$

We have to find the maximum power dissipation.

Simplify

The maximum power will be find out by putting r=R,

$P_{\max} = \frac{{rE}^{2}}{{(r + r)}^{2}}$

$P_{\max} = \frac{E^{2}}{4 r} P_{\max} = \frac{{(9 V)}^{2}}{4 \times 1 Ω} P_{\max} = \frac{81}{4} P_{\max} = 20.25 W$

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Short Answer

Step by step solution

Part (A) Step 1: Given information

Simplify

Part (B) Step 1: Given information

Simplify

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