Consider the circuit given here with the foll | vedclass

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Question

(A) The current $i$ in the circuit is given by $i = \frac{E}{R + r} = \frac{12}{4 + 2} = \frac{12}{6} = 2\, A$.$1$. Rate of energy loss in the source

Vedclass · Accepted Answer

Rate of energy loss in the source is = $8\, W$

Vedclass · Answer

(A) The current $i$ in the circuit is given by $i = \frac{E}{R + r} = \frac{12}{4 + 2} = \frac{12}{6} = 2\, A$.$1$. Rate of energy loss in the source (internal resistance) is $P_{loss} = i^2 r = (2)^2 	imes 2 = 4 	imes 2 = 8\, W$.$2$. Rate of energy conversion in the source (total power generated) is $P_{gen} = E 	imes i = 12 	imes 2 = 24\, W$.$3$. Power output across the external resistor $R$ is $P_{out} = i^2 R = (2)^2 	imes 4 = 4 	imes 4 = 16\, W$.$4$. Potential drop across $R$ is $V = iR = 2 	imes 4 = 8\, V$.Comparing these with the options,statement $(A)$ is correct.

Consider the circuit given here with the following parameters: $E.M.F.$ of the cell = $12\, V$,internal resistance of the cell = $2\,\Omega$,and external resistance $R = 4\,\Omega$. Which one of the following statements is true?

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