Six lead- acid type of secondary cells each of emf 2-0 Vand internal resistance 0-005 - are joined in series to provide a supply to a resistance of - What are the current drawn from the supply and its terminal voltage-

V_T = V ir V_T = 12 1.39 × 0.09 = 11.8749 ...

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Standard XII

Physics

Kirchhoff's Junction Law

Six lead- aci...

Question

Six lead- acid type of secondary cells each of emf $2.0 V$ and internal resistance 0.005 $Ω$ are joined in series to provide a supply to a resistance of $Ω$ . What are the current drawn from the supply and its terminal voltage?

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Solution

$V_{T} = V - i r$
$V_{T} = 12 - 1.39 \times 0.09$
$= 11.8749$

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Six lead- acid type of secondary cells each of emf 2.0Vand internal resistance 0.005 Ω are joined in series to provide a supply to a resistance of Ω. What are the current drawn from the supply and its terminal voltage?

VT=V−irVT=12−1.39×0.09=11.8749

Six lead- acid type of secondary cells each of emf $2.0 V$ and internal resistance 0.005 $Ω$ are joined in series to provide a supply to a resistance of $Ω$ . What are the current drawn from the supply and its terminal voltage?

$V_{T} = V - i r$
$V_{T} = 12 - 1.39 \times 0.09$
$= 11.8749$