The emf of a storage battery of a car is 6.0 | vedclass

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(C) The maximum power is drawn from a battery when the external resistance $R$ is equal to the internal resistance $r$ of the battery $(R = r)$.The po

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$45$

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(C) The maximum power is drawn from a battery when the external resistance $R$ is equal to the internal resistance $r$ of the battery $(R = r)$.The power delivered to the external resistor is given by the formula: $P = I^2 R = \left(\frac{E}{R+r}ight)^2 \cdot R$.Substituting $R = r$ for maximum power, we get:$P_{max} = \frac{E^2 r}{(r+r)^2} = \frac{E^2 r}{4r^2} = \frac{E^2}{4r}$.Given values are $E = 6.0 \ V$ and $r = 0.2 \ \Omega$.Substituting these values into the formula:$P_{max} = \frac{(6.0)^2}{4 	imes 0.2} = \frac{36}{0.8} = 45 \ W$.Therefore, the maximum power drawn is $45 \ W$.

The emf of a storage battery of a car is $6.0 \ V$. If the internal resistance of the battery is $0.2 \ \Omega$, then the maximum power drawn from the battery is . . . . . . $W$.

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