A 5.0 A current is set up in a circuit for 6.0 min by a rechargeable battery with a 6.0 V emf. By how much is the chemical energy of the battery reduced?

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Solution

The chemical energy of the battery is reduced by $Δ E = q ε$ , where $q$ is the charge that passes through in time $Δ t = 6.0$ min, and $ε$ is the emf of the battery. If $i$ is the current, then $q = i Δ t$ and

$Δ E = i ε Δ t = (5.0 A) (6.0 V) (6.0 m i n) (60 s / m i n) = 1.1 \times 10^{4} J$

We note the conversion of time from minutes to seconds.

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A 5.0 A current is set up in a circuit for 6.0 min by a rechargeable battery with a 6.0 V emf. By how much is the chemical energy of the battery reduced?

The chemical energy of the battery is reduced by ΔE=qε, where q is the charge that passes through in time Δt=6.0 min, and ε is the emf of the battery. If i is the current, then q=iΔt andΔE=iεΔt=(5.0A)(6.0V)(6.0min)(60s/min)=1.1×104JWe note the conversion of time from minutes to seconds.