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Liono4ka [1.6K]

3 years ago

10

A battery has an emf of 15.0 V. The terminal voltage of the battery is 12.2 V when it is delivering 14.0 W of power to an extern

al load resistor R. (a) What is the value of R?
(b) What is the internal resistance of the battery?

1 answer:

solong [7]3 years ago

3 0

Answer:

The value of R and the internal resistance of the battery are 10.6 ohm and 2.45 ohm

Explanation:

Given that,

Emf of battery = 15.0 V

Voltage = 12.2 V

Power = 14.0 W

(a). We need to calculate the value of R

Using formula of power

$P=IV$

$P=\dfrac{V^2}{R}$

$R=\dfrac{V^2}{P}$

Where, R = resistance

P = power

V = voltage

Put the value into the formula

$R =\dfrac{(12.2)^2}{14.0}$

$R=10.6\ \Omega$

(b). We need to calculate the internal resistance of the battery

Firstly we calculate the current

Using formula of current

$I=\dfrac{V}{R}=\dfrac{P}{V}$

Put the value of P and V into the formula

$I =\dfrac{14}{12.2}$

$I=1.14\ A$

We calculate the internal resistance

Using formula of emf

$e-V=Ir$

$r =\dfrac{e-V}{I}$

Put the value into the formula

$r=\dfrac{15.0-12.2}{1.14}$

$r = 2.45\ \Omega$

Hence, The value of R and the internal resistance of the battery are 10.6 ohm and 2.45 ohm

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Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is

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$a = v \frac{dv}{ds}$

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$[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.$

$\frac{200^2}{2} = c ln[\frac{3}{1.5} ]$

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