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umka2103 [35]
3 years ago
13

When a 100-Ω resistor is connected across the terminals of a battery of emf ε and internal resistance r, the battery delivers 0.

794 W of power to the 100-Ω resistor. When the 100-Ω resistor is replaced by a 200-Ω resistor, the battery delivers 0.401 W of power to the 200-Ω resistor. What are the emf and internal resistance of the battery?
Physics
1 answer:
gladu [14]3 years ago
5 0

Answer:

The emf and internal resistance of the battery are 9 volt and 2.04 Ω

Explanation:

Given that,

First resistorR = 100\ \Omega

PowerP = 0.794\ W

Second resistorR=200\ \Omega

PowerP=0.401\ W

We need to calculate the emf and internal resistance

Using formula of emf

P=\dfrac{e^2}{R+r}

For first resistor

0.794=\dfrac{e^2}{100+r}....(I)

For second resistor

0.401=\dfrac{e^2}{200+r}.....(II)

From equation (I) and (II)

\dfrac{795}{401}=\dfrac{200+r}{100+r}

Using component of dividend rule

\dfrac{393}{401}=\dfrac{100}{100+r}

r =2.04\ \Omega

Now, Put the value of internal resistance in equation (I)

0.794=\dfrac{e^2}{100+2.04}

e^2=0.794(100+2.04)

e=\sqrt{0.794(100+2.04)}

e=9\ volt

Hence, The emf and internal resistance of the battery are 9 volt and 2.04 Ω

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#2

\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\beta=2(p^2+q^2)

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\\ \rm\Rrightarrow cos\beta=0

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#3

\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\gamma=p^2+q^2

\\ \rm\Rrightarrow 2(p^2-q^2)cos\gamma=-(p^2+q^2)

\\ \rm\Rrightarrow cos\gamma =\dfrac{q^2-p^2}{2(p^2-q^2)}

\\ \rm\Rrightarrow \gamma=cos^{-1}\left(\dfrac{q^2-p^2}{2(p^2+q^2)}\right)

8 0
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We have given work done = 4780 j

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We know that kinetic energy is given  by

KE=\frac{1}{2}mv^2

So v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2\times 4780}{2.7\times 10^3}}=1.88m/sec

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