A voltmeter connected to the terminals of battery reads $6 V$ initially. When the lamp is lighted with the battery, same voltmeter reads $4 V$ . If resistance of lamp is 10 $Ω$ , resistance of battery in Ohms is

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Solution

Open circuit voltage across battery is
$ε = 6 V$

when switch is closed,
$ε - i r = 4$ and $i \times 10 = 4$

$i = 0.4 \Rightarrow 6 - 0.4 r = 4$

$\frac{6 - 4}{0.4} = r \Rightarrow r = 5 Ω$

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A voltmeter connected to the terminals of battery reads 6 V initially. When the lamp is lighted with the battery, same voltmeter reads 4 V. If resistance of lamp is 10 Ω, resistance of battery in Ohms is

Open circuit voltage across battery is ε=6 V when switch is closed, ε−ir=4 and i×10=4 i=0.4⇒6−0.4r=4 6−40.4=r⇒r=5 Ω

A voltmeter connected to the terminals of battery reads $6 V$ initially. When the lamp is lighted with the battery, same voltmeter reads $4 V$ . If resistance of lamp is 10 $Ω$ , resistance of battery in Ohms is

Open circuit voltage across battery is
$ε = 6 V$

when switch is closed,
$ε - i r = 4$ and $i \times 10 = 4$

$i = 0.4 \Rightarrow 6 - 0.4 r = 4$

$\frac{6 - 4}{0.4} = r \Rightarrow r = 5 Ω$